Electric Field Questions in English

(C) The electric field $E$ at a point outside a spherical conductor at a distance $r$ from its center is given by the formula:
$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$
Given values:
Charge $q = 3 \text{ nC} = 3 \times 10^{-9} \text{ C}$
Distance $r = 3 \text{ cm} = 3 \times 10^{-2} \text{ m}$
Constant $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \text{ Nm}^{2}\text{C}^{-2}$
Substituting these values into the formula:
$E = (9 \times 10^{9}) \times \frac{3 \times 10^{-9}}{(3 \times 10^{-2})^{2}}$
$E = \frac{9 \times 10^{9} \times 3 \times 10^{-9}}{9 \times 10^{-4}}$
$E = \frac{27}{9 \times 10^{-4}} = 3 \times 10^{4} \text{ Vm}^{-1}$
Thus,the electric field at a distance of $3 \text{ cm}$ from the center of the sphere is $3 \times 10^{4} \text{ Vm}^{-1}$.

(C) The electric field $E_1$ at a distance $r$ from an infinitely long straight wire with linear charge density $\lambda$ is given by the formula: $E_1 = \frac{\lambda}{2 \pi \epsilon_0 r} = \frac{2 k \lambda}{r}$,where $k = \frac{1}{4 \pi \epsilon_0}$.
The electric field $E_2$ at the centre of a semicircular arc of radius $r$ with linear charge density $\lambda$ is calculated by integrating the field components: $E_2 = \frac{2 k \lambda}{r}$.
Comparing the two expressions,we find that $E_1 = \frac{2 k \lambda}{r}$ and $E_2 = \frac{2 k \lambda}{r}$.
Therefore,$E_1 = E_2$.

(B) The concept of a field is a theoretical model used to explain the influence that a massive body or a charged particle extends into the space surrounding it.
This field exerts a force on any other massive body or charged particle placed within that space.
Therefore,the field concept is primarily used to understand and describe forces that act through a distance,such as gravitational and electrostatic forces,rather than contact forces.

(A) Let the charges at corners $A, B, C, D$ be $q_A = +q$,$q_B = +2q$,$q_C = +q$,and $q_D = -2q$.
At the centre $O$,the electric field due to $q_A$ and $q_C$ are equal in magnitude and opposite in direction because $q_A = q_C = +q$ and the distances $OA = OC$. Thus,they cancel each other out.
Now,consider the charges at $B$ and $D$. The charge at $B$ is $q_B = +2q$,which exerts a repulsive force on the unit positive charge at $O$ directed along $OB$ (away from $B$).
The charge at $D$ is $q_D = -2q$,which exerts an attractive force on the unit positive charge at $O$ directed along $OD$ (towards $D$).
Since both forces are directed along the diagonal $BD$ in the same direction,the net force on the unit positive charge at $O$ is directed along the diagonal $BD$.

(A) The density of electric field lines is directly proportional to the magnitude of the electric field.
Given that the separation of field lines on the left (at point $B$) is twice the separation of those on the right (at point $A$),the electric field at $B$ is half the electric field at $A$.
$E_B = \frac{E_A}{2} = \frac{40 \ Vm^{-1}}{2} = 20 \ Vm^{-1}$.
The force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$.
Given $q = 20 \ \mu C = 20 \times 10^{-6} \ C$.
$F = (20 \times 10^{-6} \ C) \times (20 \ Vm^{-1}) = 400 \times 10^{-6} \ N = 4 \times 10^{-4} \ N$.

(A) The electric field $E$ due to a point charge $q$ at a distance $r$ is given by the formula:
$E = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$
Given values are:
$E = 2 \ N \ C^{-1}$
$r = 30 \ cm = 0.3 \ m = 30 \times 10^{-2} \ m$
$\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \ m^{2} \ C^{-2}$
Substituting these values into the formula:
$2 = 9 \times 10^{9} \times \frac{q}{(30 \times 10^{-2})^{2}}$
$2 = 9 \times 10^{9} \times \frac{q}{900 \times 10^{-4}}$
$2 = 9 \times 10^{9} \times \frac{q}{9 \times 10^{-2}}$
$2 = 10^{11} \times q$
$q = \frac{2}{10^{11}} = 2 \times 10^{-11} \ C$
Thus,the magnitude of the charge is $2 \times 10^{-11} \ C$.

(A) The four charges are located at $A(1,0,0), B(-1,0,0), C(0,1,0),$ and $D(0,-1,0)$. We need to find the electric field at point $P(0,0,1)$.
The distance $r$ of point $P$ from each charge is $r = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$.
The magnitude of the electric field due to each charge at point $P$ is:
$E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2}$.
Due to symmetry,the horizontal components of the electric fields from the four charges cancel each other out. Only the vertical components (along the $Z$-axis) contribute to the net electric field.
The angle $\theta$ that the electric field vector from each charge makes with the $Z$-axis is given by $\cos \theta = \frac{1}{r} = \frac{1}{\sqrt{2}}$.
The vertical component of the electric field from each charge is $E_z = E \cos \theta = \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2} \right) \cdot \frac{1}{\sqrt{2}}$.
Since there are four such charges,the net electric field $E_{\text{net}}$ is:
$E_{\text{net}} = 4 \cdot E_z = 4 \cdot \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{2 \sqrt{2}} \right) = \frac{q}{2 \sqrt{2} \pi \varepsilon_0} \text{ N/C}$.

(D) The electric field $E$ on the surface of a charged conducting sphere is given by the formula $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space.
Since both spheres have equal surface charge densities $(\sigma_1 = \sigma_2 = \sigma)$,the electric field on the surface of both spheres depends only on the surface charge density $\sigma$ and the constant $\epsilon_0$.
Therefore,the electric field on the surface of the larger sphere is also $E = \frac{\sigma}{\epsilon_0}$.
Thus,the electric field on the surface of the larger sphere is equal to the electric field on the surface of the smaller sphere,which is $E$.

(D) Let the electric field be zero at a point $P$ located at a distance $x$ from charge $Q$ and $(6-x)$ from charge $4Q$.
At point $P$,the magnitude of the electric field due to both charges must be equal:
$\frac{KQ}{x^2} = \frac{K(4Q)}{(6-x)^2}$
$\frac{1}{x^2} = \frac{4}{(6-x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{6-x}$
$6 - x = 2x$
$3x = 6$
$x = 2 \text{ cm}$
The distance $x$ is from charge $Q$. The question asks for the distance from charge $4Q$,which is $(6-x)$.
Distance from $4Q = 6 - 2 = 4 \text{ cm}$.

(C) Let the radius of the semicircle be $R$. The length of the wire is $L = \pi R$,so $R = L / \pi$. The linear charge density is $\lambda = Q / L$.
Consider a small element of the wire of length $dl = R d\theta$ at an angle $\theta$ with the vertical axis. The charge on this element is $dQ = \lambda dl = \lambda R d\theta$.
The electric field due to this element at the centre is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dQ}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda R d\theta}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda d\theta}{R}$.
By symmetry,the horizontal components of the electric field cancel out. The net electric field is the integral of the vertical components $dE \cos \theta$ from $-\pi / 2$ to $\pi / 2$:
$E = \int_{-\pi / 2}^{\pi / 2} dE \cos \theta = \frac{\lambda}{4 \pi \varepsilon_0 R} \int_{-\pi / 2}^{\pi / 2} \cos \theta d\theta = \frac{\lambda}{4 \pi \varepsilon_0 R} [\sin \theta]_{-\pi / 2}^{\pi / 2} = \frac{\lambda}{4 \pi \varepsilon_0 R} (1 - (-1)) = \frac{2 \lambda}{4 \pi \varepsilon_0 R} = \frac{\lambda}{2 \pi \varepsilon_0 R}$.
Substituting $\lambda = Q / L$ and $R = L / \pi$:
$E = \frac{Q / L}{2 \pi \varepsilon_0 (L / \pi)} = \frac{Q}{2 \pi \varepsilon_0 L} \cdot \frac{\pi}{L} = \frac{Q}{2 \varepsilon_0 L^2} \cdot \pi = \frac{\pi Q}{2 \varepsilon_0 L^2}$.
Wait,checking the options provided,the standard result for a semicircle is $E = \frac{2k\lambda}{R} = \frac{2}{4\pi\varepsilon_0} \cdot \frac{Q/L}{L/\pi} = \frac{2\pi Q}{4\pi\varepsilon_0 L^2} = \frac{Q}{2\varepsilon_0 L^2}$.
Thus,the correct option is $C$.

(C) When the observer is at rest relative to the static charge,they only experience an electric field.
When the observer starts moving away from the charge,there is a relative velocity between the charge and the observer.
According to the principles of electromagnetism,a moving charge creates both an electric field and a magnetic field in the frame of reference of the observer.
Therefore,the observer experiences both an electric field and a magnetic field.

(D) The hexagon has charges $+Q, -Q, +Q, -Q, +Q, -Q$ at its vertices.
Since the charges are arranged in pairs of $(+Q, -Q)$ at opposite vertices,the total charge of the system is $0$.
For a system with a total charge of $0$,the electric field at a large distance $x$ is dominated by the dipole moment.
However,in this specific symmetric arrangement,the dipole moments of the three pairs of opposite charges cancel each other out because they are oriented at $120^{\circ}$ to each other.
Alternatively,considering the potential $V$ at a point on the axis at distance $x$,$V = \sum \frac{k q_i}{r_i}$. Since the charges are equal and opposite in pairs and the distance $r_i$ from any vertex to the point on the axis is the same $(r = \sqrt{x^2 + a^2})$,the potential $V = 0$ for all $x$.
Since the potential $V$ is zero everywhere on this axis,the electric field $E = -\frac{dV}{dx}$ must also be zero.

(C) The electric field due to a point charge $q$ at position vector $\vec{r}$ is given by $\vec{E} = \frac{kq}{r^3} \vec{r}$.
For point $A(1, 2, 3)$,$\vec{r}_A = \hat{i} + 2\hat{j} + 3\hat{k}$,$r_A = \sqrt{1^2+2^2+3^2} = \sqrt{14}$. So,$\vec{E}_A = \frac{kq}{14^{3/2}}(\hat{i} + 2\hat{j} + 3\hat{k})$.
For point $B(1, 1, -1)$,$\vec{r}_B = \hat{i} + \hat{j} - \hat{k}$,$r_B = \sqrt{1^2+1^2+(-1)^2} = \sqrt{3}$. So,$\vec{E}_B = \frac{kq}{3^{3/2}}(\hat{i} + \hat{j} - \hat{k})$.
For point $C(2, 2, 2)$,$\vec{r}_C = 2\hat{i} + 2\hat{j} + 2\hat{k}$,$r_C = \sqrt{2^2+2^2+2^2} = \sqrt{12} = 2\sqrt{3}$. So,$\vec{E}_C = \frac{kq}{(12)^{3/2}}(2\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{kq}{8 \cdot 3^{3/2}} \cdot 2(\hat{i} + \hat{j} + \hat{k}) = \frac{kq}{4 \cdot 3^{3/2}}(\hat{i} + \hat{j} + \hat{k})$.
Check $1$: $\vec{E}_A \cdot \vec{E}_B \propto (1)(1) + (2)(1) + (3)(-1) = 1 + 2 - 3 = 0$. Thus,$E_A \perp E_B$ is correct.
Check $3$: $|E_B| = \frac{kq}{3^{3/2}} \sqrt{1^2+1^2+(-1)^2} = \frac{kq}{3^{3/2}} \sqrt{3} = \frac{kq}{3}$.
$|E_C| = \frac{kq}{4 \cdot 3^{3/2}} \sqrt{1^2+1^2+1^2} = \frac{kq \sqrt{3}}{4 \cdot 3^{3/2}} = \frac{kq}{4 \cdot 3} = \frac{kq}{12}$.
Therefore,$|E_B| = 4|E_C|$ is correct.

(A) Let the charges be $q_A = -5 \mu C$ and $q_B = +5 \mu C$. Let the distance $AB = d = 5 \ cm$. Let point $C$ be at a distance $x$ from $A$ and $y$ from $B$. For the resultant electric field at $C$ to be parallel to the line $AB$,the vertical components of the electric fields produced by $q_A$ and $q_B$ must cancel each other out.
Let $\theta_A$ and $\theta_B$ be the angles made by the vectors $\vec{E}_A$ and $\vec{E}_B$ with the line $AB$. For the vertical components to cancel,we must have $E_A \sin \theta_A = E_B \sin \theta_B$.
Since the magnitudes of the charges are equal $(|q_A| = |q_B| = q)$,the field magnitudes are $E_A = \frac{kq}{x^2}$ and $E_B = \frac{kq}{y^2}$.
Thus,$\frac{kq}{x^2} \sin \theta_A = \frac{kq}{y^2} \sin \theta_B$. From the geometry of the triangle $ABC$,$\sin \theta_A = \frac{h}{x}$ and $\sin \theta_B = \frac{h}{y}$,where $h$ is the perpendicular distance of $C$ from line $AB$.
Substituting these,we get $\frac{kq}{x^2} \cdot \frac{h}{x} = \frac{kq}{y^2} \cdot \frac{h}{y}$,which simplifies to $\frac{1}{x^3} = \frac{1}{y^3}$,implying $x = y$.
Therefore,$AC = BC$.

(B) Let the square be $ABCD$ with side length $l$. Let the charges be at corners $A, B, C, D$. Consider the midpoint $M$ of side $CD$. The electric fields due to charges at $C$ and $D$ are $E_C$ and $E_D$. Since $MC = MD = l/2$,$E_C = k q / (l/2)^2$ directed away from $C$,and $E_D = k q / (l/2)^2$ directed away from $D$. These are equal and opposite,so they cancel out.
Now,consider the fields due to charges at $A$ and $B$. The distance from $A$ to $M$ is $r = \sqrt{l^2 + (l/2)^2} = \sqrt{5l^2/4} = l\sqrt{5}/2$. The magnitude of the field is $E_A = E_B = k q / r^2 = k q / (5l^2/4) = 4kq / 5l^2$.
The vertical components of $E_A$ and $E_B$ add up,while horizontal components cancel. Let $\theta$ be the angle the vector $AM$ makes with the vertical. Then $\cos \theta = l / r = l / (l\sqrt{5}/2) = 2/\sqrt{5}$.
The net electric field is $E_{net} = 2 E_A \cos \theta = 2 \times (4kq / 5l^2) \times (2/\sqrt{5}) = 16kq / 5\sqrt{5}l^2$.

(B) The electric field $E$ due to an infinitely long wire with linear charge density $\lambda$ at a distance $r$ is given by $E = \frac{\lambda}{2 \pi \epsilon_0 r} = \frac{2k\lambda}{r}$,where $k = 9 \times 10^9 \ Nm^2C^{-2}$.
Given $\lambda_1 = 4 \ Cm^{-1}$,$\lambda_2 = 8 \ Cm^{-1}$,and separation $d = 4 \ cm = 0.04 \ m$.
The midpoint is at distance $r = d/2 = 0.02 \ m$ from both wires.
The electric field due to wire $1$ is $E_1 = \frac{2 \times 9 \times 10^9 \times 4}{0.02} = 36 \times 10^{11} \ NC^{-1}$ (directed away from wire $1$).
The electric field due to wire $2$ is $E_2 = \frac{2 \times 9 \times 10^9 \times 8}{0.02} = 72 \times 10^{11} \ NC^{-1}$ (directed away from wire $2$).
Since both wires have positive charge densities,the fields at the midpoint are in opposite directions.
The net electric field is $E_{net} = |E_2 - E_1| = |72 \times 10^{11} - 36 \times 10^{11}| = 36 \times 10^{11} \ NC^{-1}$.

(A) Given,electric field $E = 500 \ Vm^{-1}$ directed at $\theta = 30^{\circ}$ with the positive $X$-axis.
The electric field vector is $\vec{E} = E(\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j}) = 500 \left(\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}\right) = (250\sqrt{3} \hat{i} + 250 \hat{j}) \ Vm^{-1}$.
The coordinates of point $A$ are $(-3, 0) \ m$ and point $B$ are $(0, 5) \ m$.
The displacement vector from $A$ to $B$ is $\vec{r}_{AB} = \vec{r}_B - \vec{r}_A = (0 - (-3)) \hat{i} + (5 - 0) \hat{j} = (3 \hat{i} + 5 \hat{j}) \ m$.
The potential difference is given by $\Delta V = V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r} = -\vec{E} \cdot \vec{r}_{AB}$.
Substituting the values:
$V_B - V_A = -(250\sqrt{3} \hat{i} + 250 \hat{j}) \cdot (3 \hat{i} + 5 \hat{j})$
$V_B - V_A = -(250\sqrt{3} \times 3 + 250 \times 5)$
$V_B - V_A = -250(3\sqrt{3} + 5) \ V$.

(A) Given: Charge $q = 5 C$,Force $F = 5000 N$,Distance $d = 1 cm = 10^{-2} m$.
First,calculate the electric field intensity $E$ using the formula $E = F / q$.
$E = 5000 / 5 = 1000 N/C$.
The potential difference $V$ between two points separated by distance $d$ in a uniform electric field is given by $V = E \times d$.
Substituting the values: $V = 1000 \times 10^{-2} = 10 V$.

(D) The potential in an electric field is given by $V = (x^2 - y^2)$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = 2x$ and $\frac{\partial V}{\partial y} = -2y$.
Thus, $\vec{E} = -(2x \hat{i} - 2y \hat{j}) = -2x \hat{i} + 2y \hat{j}$.
The differential equation for the electric field lines is given by $\frac{dy}{dx} = \frac{E_y}{E_x}$.
Substituting the components of $\vec{E}$:
$\frac{dy}{dx} = \frac{2y}{-2x} = -\frac{y}{x}$.
Rearranging the terms, we get $\frac{dy}{y} = -\frac{dx}{x}$.
Integrating both sides, $\ln y = -\ln x + C$, which simplifies to $\ln(xy) = C$, or $xy = \text{constant}$.
This equation represents a rectangular hyperbola. Among the given options, the graph of rectangular hyperbolas is shown in option $(d)$.

(A) The electric field $E$ inside a uniformly charged solid sphere at a distance $r$ from the center is given by $E = \frac{Q r}{4 \pi \varepsilon_0 R^3}$.
Since the particle has a charge $-q$,the restoring force acting on it is $F = -qE = -\frac{Q q r}{4 \pi \varepsilon_0 R^3}$.
This force is of the form $F = -kr$,where the force constant $k = \frac{Q q}{4 \pi \varepsilon_0 R^3}$.
The frequency of oscillation $f$ is given by $f = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.
Substituting the value of $k$,we get $f = \frac{1}{2 \pi} \sqrt{\frac{Q q}{4 \pi \varepsilon_0 R^3 m}}$.

(D) Let the total charge on the loop be $Q = 10^{-5} \ C$ and the radius be $a = 10 \ cm = 0.1 \ m$.
The linear charge density is $\lambda = \frac{Q}{2 \pi a}$.
The charge on the small cut-off element of length $dl = 3.14 \times 10^{-6} \ m$ is $dq = \lambda dl = \frac{Q dl}{2 \pi a}$.
Initially,the electric field at the center due to the complete loop is zero.
If $E_{\text{rem}}$ is the field due to the remaining wire and $E_{dl}$ is the field due to the cut-off element,then $E_{\text{rem}} + E_{dl} = 0$,which implies $|E_{\text{rem}}| = |E_{dl}|$.
The electric field due to the small element $dq$ at the center is $E_{dl} = \frac{k dq}{a^2} = \frac{k Q dl}{a^2 (2 \pi a)} = \frac{k Q dl}{2 \pi a^3}$.
Substituting the values:
$E_{dl} = \frac{(9 \times 10^9) \times (10^{-5}) \times (3.14 \times 10^{-6})}{2 \times \pi \times (0.1)^3}$
Since $2 \pi \approx 2 \times 3.14 = 6.28$,we have $3.14 / (2 \pi) = 0.5$.
$E_{dl} = \frac{9 \times 10^9 \times 10^{-5} \times 10^{-6} \times 0.5}{0.001} = \frac{9 \times 10^{-2} \times 0.5}{10^{-3}} = 4.5 \times 10^1 = 45 \ N \ C^{-1}$.

(C) Consider an infinitesimal section of the rod of length $dx$,at a distance $x$ from the origin,as shown in the figure.
It contains charge $dq = \lambda dx$ and is at a distance $r = \sqrt{x^2 + L^2}$ from the point $(0, L)$.
The magnitude of the electric field produced by $dx$ at the point $(0, L)$ is $dE = \frac{k dq}{r^2} = \frac{k \lambda dx}{x^2 + L^2}$.
The $x$ and $y$ components of the electric field are $dE_x = dE \sin \theta$ and $dE_y = dE \cos \theta$,where $\tan \theta = \frac{x}{L}$.
Thus,$x = L \tan \theta$ and $dx = L \sec^2 \theta d\theta$. Also,$r^2 = L^2 \sec^2 \theta$.
Substituting these into the components:
$dE_x = \frac{k \lambda (L \sec^2 \theta d\theta)}{L^2 \sec^2 \theta} \sin \theta = \frac{k \lambda}{L} \sin \theta d\theta$.
$dE_y = \frac{k \lambda (L \sec^2 \theta d\theta)}{L^2 \sec^2 \theta} \cos \theta = \frac{k \lambda}{L} \cos \theta d\theta$.
Integrating from $x=0$ to $x=\infty$ corresponds to $\theta$ from $0$ to $\frac{\pi}{2}$:
$E_x = \int_0^{\pi/2} \frac{k \lambda}{L} \sin \theta d\theta = \frac{k \lambda}{L} [-\cos \theta]_0^{\pi/2} = \frac{k \lambda}{L}$.
$E_y = \int_0^{\pi/2} \frac{k \lambda}{L} \cos \theta d\theta = \frac{k \lambda}{L} [\sin \theta]_0^{\pi/2} = \frac{k \lambda}{L}$.
The magnitude of the resultant electric field is $E = \sqrt{E_x^2 + E_y^2} = \sqrt{(\frac{k \lambda}{L})^2 + (\frac{k \lambda}{L})^2} = \sqrt{2} \frac{k \lambda}{L}$.
Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $E = \frac{\sqrt{2} \lambda}{4 \pi \varepsilon_0 L} = \frac{\lambda}{2 \sqrt{2} \pi \varepsilon_0 L}$.

(A) Let the point where the resultant electric field is zero be at a distance $x$ from the $-8 \mu C$ charge. Since the charges have opposite signs,the null point must lie outside the region between the charges,specifically on the side of the smaller magnitude charge $(-8 \mu C)$.
Let the distance from $-8 \mu C$ be $x$. Then the distance from $+32 \mu C$ is $(15 + x)$.
At the null point,the magnitudes of the electric fields produced by both charges must be equal:
$\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(15 + x)^2}$
$\frac{8 \times 10^{-6}}{x^2} = \frac{32 \times 10^{-6}}{(15 + x)^2}$
$\frac{1}{x^2} = \frac{4}{(15 + x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{15 + x}$
$15 + x = 2x$
$x = 15 \ cm$.

(C) Let the charges be $q_1 = -10 \mu C$ at $x_1 = 0$ and $q_2 = +5 \mu C$ at $x_2 = \sqrt{2} \ m$.
Since the charges have opposite signs,the electric field can only be zero at a point outside the line segment joining the charges,specifically on the side of the smaller magnitude charge $(q_2)$.
Let the neutral point be at distance $d$ from $q_2$ in the positive $x$-direction.
The electric field due to $q_1$ and $q_2$ at this point must be equal in magnitude:
$\frac{k|q_1|}{(d + \sqrt{2})^2} = \frac{k|q_2|}{d^2}$
$\frac{10}{(d + \sqrt{2})^2} = \frac{5}{d^2}$
$2d^2 = (d + \sqrt{2})^2$
Taking the square root on both sides:
$\sqrt{2}d = d + \sqrt{2}$
$d(\sqrt{2} - 1) = \sqrt{2}$
$d = \frac{\sqrt{2}}{\sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} + 1)}{2 - 1} = 2 + \sqrt{2} \ m$.
The coordinate $x$ of this point is $x_2 + d = \sqrt{2} + (2 + \sqrt{2}) = 2 + 2\sqrt{2} = 2(\sqrt{2} + 1) \ m$.

(D) The electric field $E$ near the surface of a conducting sphere with surface charge density $\sigma$ is given by the formula $E = \frac{\sigma}{\epsilon_0}$.
Since both spheres are charged to the same surface charge density,we have $\sigma_1 = \sigma_2 = \sigma$.
Therefore,the electric field near the surface of the first sphere is $E_1 = \frac{\sigma}{\epsilon_0}$.
Similarly,the electric field near the surface of the second sphere is $E_2 = \frac{\sigma}{\epsilon_0}$.
Comparing the two,we find $E_1 = E_2$.
Thus,the ratio of the electric fields is $\frac{E_1}{E_2} = \frac{1}{1}$.

(B) The electric field $E$ due to an infinitely long straight wire with linear charge density $\lambda$ at a distance $r$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.
For the first wire with charge density $\lambda_1 = 2 \lambda$,the electric field at the midpoint (distance $r = R/2$) is $E_1 = \frac{2 \lambda}{2 \pi \varepsilon_0 (R/2)} = \frac{2 \lambda}{\pi \varepsilon_0 R}$.
For the second wire with charge density $\lambda_2 = 3 \lambda$,the electric field at the midpoint (distance $r = R/2$) is $E_2 = \frac{3 \lambda}{2 \pi \varepsilon_0 (R/2)} = \frac{3 \lambda}{\pi \varepsilon_0 R}$.
Since the wires are parallel and positively charged,the fields at the midpoint point in opposite directions.
The net electric field intensity is $E_{net} = |E_2 - E_1| = |\frac{3 \lambda}{\pi \varepsilon_0 R} - \frac{2 \lambda}{\pi \varepsilon_0 R}| = \frac{\lambda}{\pi \varepsilon_0 R}$.

(C) Given,linear charge density $\lambda = \lambda_0 \cos \phi$.
Consider two symmetric elements of length $dl = r d\phi$ at an angle $\phi$ on both sides of the vertical axis.
The charge on each element is $dq = \lambda dl = (\lambda_0 \cos \phi) r d\phi$.
The electric field due to each element at the centre is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{r^2} = \frac{\lambda_0 \cos \phi d\phi}{4 \pi \varepsilon_0 r}$.
The horizontal components $dE \sin \phi$ cancel out due to symmetry.
The net electric field is the sum of vertical components $dE \cos \phi$:
$E = \int_0^{\pi} 2 (dE \cos \phi) = \int_0^{\pi} 2 \left( \frac{\lambda_0 \cos \phi d\phi}{4 \pi \varepsilon_0 r} \right) \cos \phi$
$E = \frac{2 \lambda_0}{4 \pi \varepsilon_0 r} \int_0^{\pi} \cos^2 \phi d\phi$
Using $\cos^2 \phi = \frac{1 + \cos 2\phi}{2}$:
$E = \frac{\lambda_0}{4 \pi \varepsilon_0 r} \int_0^{\pi} (1 + \cos 2\phi) d\phi = \frac{\lambda_0}{4 \pi \varepsilon_0 r} [\phi + \frac{\sin 2\phi}{2}]_0^{\pi} = \frac{\lambda_0}{4 \pi \varepsilon_0 r} [\pi - 0] = \frac{\lambda_0}{4 \varepsilon_0 r}$.

(C) Radius of the loop, $r = 1 \, cm = 0.01 \, m$.
Total charge on the loop, $Q = 1 \times 10^{-6} \, C$.
Charge per unit length, $\lambda = \frac{Q}{2 \pi r} = \frac{10^{-6}}{2 \pi \times 0.01} = \frac{10^{-4}}{2 \pi} \, C/m$.
The length of the cut-off part is $l' = 0.01 \% \, \text{of} \, 2 \pi r = \frac{0.01}{100} \times 2 \pi r = 2 \pi \times 10^{-6} \, m$.
The charge on the cut-off part is $Q' = l' \lambda = (2 \pi \times 10^{-6}) \times \frac{10^{-4}}{2 \pi} = 10^{-10} \, C$.
For a complete loop, the electric field at the centre is zero. If a small part $l'$ is removed, the electric field at the centre due to the remaining part is equal in magnitude to the electric field that would have been produced by the removed part $l'$ at the centre.
Treating the small cut-off part as a point charge $Q'$ at distance $r$ from the centre, the electric field is $E = \frac{k Q'}{r^2}$.
$E = \frac{9 \times 10^9 \times 10^{-10}}{(0.01)^2} = \frac{0.9}{10^{-4}} = 9000 \, N/C = 9 \times 10^3 \, N/C$.

(B) The electric field due to a negative charge is directed towards the charge.
Let the two negative charges be $-Q$ located at $(0, a)$ and $(0, -a)$. Point $P$ is on the positive $x$-axis at $(x, 0)$.
The electric field vector $\vec{E}_1$ due to the charge at $(0, a)$ points from $P$ towards $(0, a)$.
The electric field vector $\vec{E}_2$ due to the charge at $(0, -a)$ points from $P$ towards $(0, -a)$.
Since the magnitudes of the charges are equal and their distances from $P$ are equal,the magnitudes of the electric fields are equal,i.e.,$|\vec{E}_1| = |\vec{E}_2|$.
When resolving these vectors into components,the $y$-components ($E_{1y}$ and $E_{2y}$) are equal in magnitude but opposite in direction,so they cancel each other out.
The $x$-components ($E_{1x}$ and $E_{2x}$) both point in the negative $x$-direction.
Therefore,the resultant electric field at point $P$ is directed along the negative $x$-direction.

(B) Let the centre of the cube be the origin $(0,0,0)$. The vertices of the cube are at $(\pm L/2, \pm L/2, \pm L/2)$.
If all eight vertices had a charge $+q$,the electric field at the centre would be zero due to symmetry.
Let the charge at one vertex (say $(-L/2, -L/2, -L/2)$) be $-q$ instead of $+q$.
Let $\vec{E}_{total}$ be the electric field at the centre with the given configuration.
Let $\vec{E}_{all}$ be the electric field if all eight vertices had charge $+q$,so $\vec{E}_{all} = 0$.
Let $\vec{E}_{vertex}$ be the electric field at the centre due to a charge $+q$ at the vertex $(-L/2, -L/2, -L/2)$.
Then,$\vec{E}_{total} = \vec{E}_{all} - \vec{E}_{vertex} + \vec{E}_{(-q)} = 0 - \vec{E}_{vertex} - \vec{E}_{vertex} = -2\vec{E}_{vertex}$.
The distance $r$ from any vertex to the centre is $r = \sqrt{(L/2)^2 + (L/2)^2 + (L/2)^2} = \sqrt{3L^2/4} = \frac{\sqrt{3}L}{2}$.
The magnitude of the electric field due to one charge $q$ at the centre is $|E_{vertex}| = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q}{3L^2/4} = \frac{4}{3} \left(\frac{q}{4\pi\varepsilon_0 L^2}\right)$.
Since $\vec{E}_{total} = -2\vec{E}_{vertex}$,the magnitude is $|E_{total}| = 2 |E_{vertex}| = 2 \times \frac{4}{3} \left(\frac{q}{4\pi\varepsilon_0 L^2}\right) = \frac{8}{3} \left(\frac{q}{4\pi\varepsilon_0 L^2}\right)$.
Comparing this with $|E| = \alpha \left(\frac{q}{4\pi\varepsilon_0 L^2}\right)$,we get $\alpha = \frac{8}{3}$.

(A) When a negative charge $(-q)$ is placed at the centre of a non-conducting sphere,it creates an electric field in the surrounding space.
By definition,the electric field lines for a negative point charge are directed towards the charge.
Since the charge is at the centre of the sphere,the electric field lines at any point on the surface of the sphere will point towards the centre.
Therefore,the direction of the electric field at any point on the surface is radially inward,as shown in the figure.

(A) The separation vector $\vec{r}_{sep} = \vec{r} - \vec{r}_0 = (8 \hat{i} - 5 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = 6 \hat{i} - 8 \hat{j} \ m$.
The distance between the points is $r = |\vec{r}_{sep}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ m$.
The magnitude of the electric field is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{|q|}{r^2}$.
Substituting the values: $E = (9 \times 10^9) \times \frac{50 \times 10^{-6}}{10^2}$.
$E = (9 \times 10^9) \times \frac{50 \times 10^{-6}}{100} = 9 \times 10^9 \times 0.5 \times 10^{-6} = 4.5 \times 10^3 \ N \ C^{-1}$.
Since $1 \ N \ C^{-1} = 1 \ V \ m^{-1}$,we have $E = 4.5 \times 10^3 \ V \ m^{-1} = 4.5 \ kV \ m^{-1}$.

(C) The electric field intensity due to a point charge is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.
Since the consecutive charges are of opposite signs, the net electric field at $x = 0$ is the sum of fields due to individual charges:
$E = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{Q}{r_1^2} - \frac{Q}{r_2^2} + \frac{Q}{r_3^2} - \frac{Q}{r_4^2} + \dots \infty \right]$
$E = \frac{Q}{4 \pi \varepsilon_0} \left[ \frac{1}{(1 \times 10^{-2})^2} - \frac{1}{(2 \times 10^{-2})^2} + \frac{1}{(4 \times 10^{-2})^2} - \frac{1}{(8 \times 10^{-2})^2} + \dots \infty \right]$
$E = (9 \times 10^9) \times (5 \times 10^{-9}) \times 10^4 \left[ \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \dots \infty \right]$
$E = 45 \times 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots \infty \right]$
This is an infinite geometric progression with first term $a = 1$ and common ratio $r = -\frac{1}{4}$.
The sum $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.
Therefore, $E = 45 \times 10^4 \times \frac{4}{5} = 36 \times 10^4 \text{ N/C}$.

(B) Let the charge $q$ be displaced by a small distance $y$ along the $y$-axis. The position of the charge becomes $(0, y)$.
The distance $r$ of this charge from each of the charges $-q$ located at $(-a, 0)$ and $(a, 0)$ is $r = \sqrt{a^2 + y^2}$.
The force exerted by each charge $-q$ on $q$ is $F' = \frac{kq^2}{r^2} = \frac{kq^2}{a^2 + y^2}$.
The components of these forces along the $x$-axis cancel out due to symmetry.
The components along the $y$-axis are both directed towards the origin (negative $y$-direction).
The net force is $F = -2 F' \sin \theta$,where $\sin \theta = \frac{y}{r} = \frac{y}{\sqrt{a^2 + y^2}}$.
$F = -2 \left( \frac{kq^2}{a^2 + y^2} \right) \left( \frac{y}{\sqrt{a^2 + y^2}} \right) = -\frac{2kq^2 y}{(a^2 + y^2)^{3/2}}$.
Since the displacement $y$ is very small $(y \ll a)$,we can approximate $(a^2 + y^2)^{3/2} \approx (a^2)^{3/2} = a^3$.
Thus,$F \approx -\frac{2kq^2}{a^3} y$.
Therefore,$F \propto -y$.

(D) The linear charge density is given by $\lambda = \lambda_0 \sin \theta$. Consider a small element of the rod subtending an angle $d\theta$ at the centre. The charge on this element is $dq = \lambda (R d\theta) = \lambda_0 R \sin \theta d\theta$.
The electric field $dE$ at the centre due to this element is $dE = \frac{k dq}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda_0 R \sin \theta d\theta}{R^2} = \frac{\lambda_0}{4 \pi \varepsilon_0 R} \sin \theta d\theta$.
Due to symmetry,the $x$-components of the electric field cancel out. The $y$-component is $dE_y = -dE \sin \theta = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \sin^2 \theta d\theta$.
Integrating from $\theta = 0$ to $\pi$:
$E_y = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \int_0^{\pi} \sin^2 \theta d\theta = -\frac{\lambda_0}{4 \pi \varepsilon_0 R} \int_0^{\pi} \frac{1 - \cos 2\theta}{2} d\theta = -\frac{\lambda_0}{8 \pi \varepsilon_0 R} [\theta - \frac{\sin 2\theta}{2}]_0^{\pi} = -\frac{\lambda_0}{8 \pi \varepsilon_0 R} (\pi) = -\frac{\lambda_0}{8 \varepsilon_0 R}$.
Thus,$\vec{E} = -\frac{\lambda_0}{8 \varepsilon_0 R} \hat{j}$. Since this is not among the options,the correct answer is $D$.

(D) The electric field $E_z$ on the axis of a uniformly charged ring of radius '$a$' at a distance '$z$' from its center is given by:
$E_z = \frac{1}{4\pi\epsilon_0} \frac{Qz}{(z^2 + a^2)^{3/2}}$
To find the position where the electric field is maximum,we differentiate $E_z$ with respect to '$z$' and set it to zero:
$\frac{dE_z}{dz} = 0$
$\frac{d}{dz} \left[ \frac{Qz}{(z^2 + a^2)^{3/2}} \right] = 0$
Using the quotient rule:
$(z^2 + a^2)^{3/2} - z \cdot \frac{3}{2}(z^2 + a^2)^{1/2} \cdot 2z = 0$
$(z^2 + a^2)^{3/2} = 3z^2(z^2 + a^2)^{1/2}$
$z^2 + a^2 = 3z^2$
$2z^2 = a^2$
$z = \frac{a}{\sqrt{2}}$
Since the horizontal axis of the graph represents $Z/a$,the coordinate $M$ corresponds to $z/a = 1/\sqrt{2}$.

(A) If there were $12$ equal charges $+Q$ at all hour positions,the electric field at the centre would be zero due to symmetry,as each charge would be cancelled by an equal and opposite charge at the diametrically opposite position.
Let $\vec{E}_{10}$ be the electric field at the centre due to a charge $+Q$ at the $10$ o'clock position. The net electric field $\vec{E}_{net}$ with the charge at $10$ missing is given by:
$\vec{E}_{total} = \vec{E}_{net} + \vec{E}_{10} = 0$
Therefore,$\vec{E}_{net} = -\vec{E}_{10}$.
The magnitude of the electric field due to a single charge $+Q$ at distance $r$ is $E = \frac{Q}{4 \pi \varepsilon_{0} r^{2}}$.
The direction of $\vec{E}_{10}$ is from the $10$ o'clock position towards the centre.
Thus,$-\vec{E}_{10}$ is a vector of magnitude $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ directed from the centre towards the $10$ o'clock position.

(A) Let the two positive charges $Q$ be located at $(0, d)$ and $(0, -d)$. The negative charge $-q$ is displaced to $(x, 0)$.
The distance $r$ between each positive charge $Q$ and the negative charge $-q$ is $r = \sqrt{x^2 + d^2}$.
The magnitude of the electrostatic force exerted by each positive charge on the negative charge is $F_e = \frac{kQq}{r^2}$.
The components of these forces along the $y$-axis cancel out due to symmetry. The components along the $x$-axis add up:
$F = -2 F_e \cos \theta$,where $\cos \theta = \frac{x}{r}$.
Substituting the expressions:
$F = -2 \left( \frac{kQq}{r^2} \right) \left( \frac{x}{r} \right) = -\frac{2kQqx}{r^3}$.
Since $r = (x^2 + d^2)^{1/2}$,we have:
$F = -\frac{2kQqx}{(x^2 + d^2)^{3/2}}$.
Given the condition $x \ll d$,we can approximate $x^2 + d^2 \approx d^2$ in the denominator:
$F \approx -\frac{2kQqx}{(d^2)^{3/2}} = -\frac{2kQqx}{d^3}$.
Thus,the magnitude of the force $F$ is directly proportional to the displacement $x$,i.e.,$F \propto x$.

(C) In a regular hexagon,the distance from the centre $O$ to any vertex is equal to the side length $a$.
To maximize the electric field at the centre,we place the four charges at vertices $A, B, C,$ and $F$ as shown in the figure.
The electric field due to charges at $F$ and $C$ are equal in magnitude and opposite in direction,so they cancel each other out $(E_F + E_C = 0)$.
The net electric field at the centre $O$ is due to the charges at $A$ and $B$.
The magnitude of the electric field due to each charge is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{a^2}$.
The angle between the electric field vectors $E_A$ and $E_B$ is $60^{\circ}$.
The resultant electric field $E_{\text{net}}$ is given by:
$E_{\text{net}} = \sqrt{E_A^2 + E_B^2 + 2 E_A E_B \cos 60^{\circ}}$
Since $E_A = E_B = E$,we have:
$E_{\text{net}} = \sqrt{E^2 + E^2 + 2 E^2 (1/2)} = \sqrt{3E^2} = E\sqrt{3}$
Substituting the value of $E$:
$E_{\text{net}} = \frac{\sqrt{3}Q}{4 \pi \epsilon_0 a^2}$

(C) Let the unit negative charge be at a distance $x$ from the midpoint along the perpendicular bisector. The distance of this charge from each fixed charge $+Q$ is $r = \sqrt{x^2 + a^2}$.
The electrostatic force exerted by each $+Q$ charge on the unit negative charge is $F = \frac{1}{4 \pi \varepsilon_0} \frac{Q \cdot 1}{r^2} = \frac{Q}{4 \pi \varepsilon_0 (x^2 + a^2)}$.
The components of these forces perpendicular to the line connecting the charges cancel out,while the components along the perpendicular bisector add up.
The net restoring force is $F_{\text{net}} = -2F \cos \theta$,where $\cos \theta = \frac{x}{r} = \frac{x}{\sqrt{x^2 + a^2}}$.
$F_{\text{net}} = -2 \left( \frac{Q}{4 \pi \varepsilon_0 (x^2 + a^2)} \right) \left( \frac{x}{\sqrt{x^2 + a^2}} \right) = -\frac{2Qx}{4 \pi \varepsilon_0 (x^2 + a^2)^{3/2}}$.
Since $x \ll a$,we can approximate $(x^2 + a^2)^{3/2} \approx (a^2)^{3/2} = a^3$.
Thus,$F_{\text{net}} \approx -\left( \frac{2Q}{4 \pi \varepsilon_0 a^3} \right) x = -\left( \frac{Q}{2 \pi \varepsilon_0 a^3} \right) x$.
This is the equation of simple harmonic motion $F = -kx_{eff}$,where $k_{eff} = \frac{Q}{2 \pi \varepsilon_0 a^3}$.
The frequency of oscillation is $f = \frac{1}{2 \pi} \sqrt{\frac{k_{eff}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{Q}{2 \pi \varepsilon_0 M a^3}}$.
Comparing this with the given options,none of the options match the calculated frequency.

(A) $1$. Electric potential $(V)$ is a scalar quantity. The potential at the center $O$ due to a single charge $q$ at a distance $r$ is $V_i = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$. Since there are five such charges,the total potential is the algebraic sum: $V = 5 \times \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} = \frac{5 q}{4 \pi \varepsilon_0 r}$.
$2$. Electric field $(\vec{E})$ is a vector quantity. For a regular polygon with identical charges at all vertices,the electric field vectors due to individual charges cancel each other out due to symmetry. Therefore,the net electric field at the center $O$ is $\vec{E} = 0$.

(B) Let the magnitude of the electric field due to a single charge $Q$ at the center be $E_{0} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{R^{2}}$.
By analyzing the symmetry,the charges at $90^{\circ}$ and $270^{\circ}$ (both $+Q$) cancel each other out.
The remaining charges are at $30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}$.
Specifically,the charges at $30^{\circ}$ and $210^{\circ}$ are $+Q$ and $+Q$,and at $150^{\circ}$ and $330^{\circ}$ are $-Q$ and $+Q$.
Using the principle of superposition,the net electric field is the vector sum of individual fields.
After resolving the components,the net electric field at the center is calculated as $\vec{E}_{net} = -\frac{Q}{4\pi\epsilon_{0}R^{2}}(\sqrt{3}\hat{i}-\hat{j})$.

(B) For a point charge $2q$ at distance $r$,the electric field is $E = k \frac{2q}{r^2}$.
For a thin spherical shell of charge $q$ and radius $R$,the electric field at a distance $r' = \frac{r}{2}$ from the center is considered. Since $r \gg R$,the condition $r' > R$ is satisfied,meaning the shell acts as a point charge at its center.
Thus,$E' = k \frac{q}{(r/2)^2} = k \frac{q}{r^2/4} = 4k \frac{q}{r^2}$.
From the first equation,$k \frac{q}{r^2} = \frac{E}{2}$.
Substituting this into the expression for $E'$,we get $E' = 4 \times \frac{E}{2} = 2E$.
Therefore,the correct option is $B$.

(A) For a semi-circular ring of radius $R$ and total charge $Q$,the linear charge density is $\lambda = \frac{Q}{\pi R}$.
The electric field $E$ at the center is given by the formula $E = \frac{2k\lambda}{R}$,where $k = \frac{1}{4\pi\epsilon_0}$.
Substituting $\lambda$,we get $E = \frac{2(1/4\pi\epsilon_0)(Q/\pi R)}{R} = \frac{Q}{2\pi^2 \epsilon_0 R^2}$.
Given $E = 100 \text{ V/m}$,$R = 0.35 \text{ m}$,$\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$,and $\pi = 3.14$.
Rearranging for $Q$: $Q = E \times 2\pi^2 \epsilon_0 R^2$.
$Q = 100 \times 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2$.
$Q = 200 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225$.
$Q \approx 2.14 \times 10^{-9} \text{ C} = 2.14 \text{ nC}$.