Relation between Electric Field and Potential and Potential Gradient Questions in English

(A) The electric potential is given by $V(x, y, z) = 4x^2 \text{ volt}$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$
$\frac{\partial V}{\partial y} = 0$
$\frac{\partial V}{\partial z} = 0$
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -(8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} \text{ V/m}$.
At the point $(1 \text{ m}, 0, 2 \text{ m})$,the $x$-coordinate is $1 \text{ m}$.
Therefore,$\vec{E} = -8(1) \hat{i} = -8 \hat{i} \text{ V/m}$.
The negative sign indicates that the electric field is directed along the negative $X$-axis with a magnitude of $8 \text{ V/m}$.

(B) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.
Given that the electric field is uniform and directed along the positive $X$-axis,we have $E = E_0$.
Substituting this into the formula: $E_0 = -\frac{dV}{dx}$.
Rearranging the terms to integrate: $dV = -E_0 dx$.
Integrating both sides from the reference point $x = 0$ (where $V = 0$) to a point $x$ (where potential is $V_x$):
$\int_{0}^{V_x} dV = -\int_{0}^{x} E_0 dx$.
$[V]_{0}^{V_x} = -E_0 [x]_{0}^{x}$.
$V_x - 0 = -E_0 (x - 0)$.
$V_x = -x E_0$.
Therefore,the potential at $x$ is $-x E_0$.

(B) The electric field $E$ between two parallel plates with a potential difference $V$ and separation distance $d$ is given by the formula $E = \frac{V}{d}$.
Given:
Potential difference $V = 10\,V$
Distance $d = 2\,cm = 2 \times 10^{-2}\,m$
Substituting the values:
$E = \frac{10}{2 \times 10^{-2}} = 5 \times 10^2 = 500\,N/C$.
Therefore,the correct option is $B$.

(A) Given: Charge $Q = 5\,C$,Force $F = 5000\,N$,Distance $d = 1\,cm = 10^{-2}\,m$.
We know that the electric field $E = \frac{F}{Q} = \frac{5000}{5} = 1000\,N/C$.
The relationship between electric field $E$,potential difference $V$,and distance $d$ is $E = \frac{V}{d}$.
Therefore,$V = E \times d$.
Substituting the values: $V = 1000 \times 10^{-2} = 10\,V$.
Thus,the potential difference is $10\,V$.

(A) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.
Given $V = 5x^2 + 10x - 9$.
Now,differentiate $V$ with respect to $x$:
$\frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) = 10x + 10$.
Substituting this into the electric field formula:
$E = -(10x + 10) = -10x - 10$.
To find the electric field at $x = 1 \text{ m}$,substitute $x = 1$ into the expression:
$E = -10(1) - 10 = -10 - 10 = -20 \text{ V/m}$.

(A) The electric field $E$ between two parallel plates is given by the formula $E = \frac{\Delta V}{d}$,where $\Delta V$ is the potential difference and $d$ is the separation distance.
Given: Potential difference $\Delta V = V_2 - V_1 = 30\, V - (-10\, V) = 40\, V$.
Separation $d = 2\, cm = 2 \times 10^{-2}\, m$.
Substituting these values into the formula:
$E = \frac{40}{2 \times 10^{-2}} = 20 \times 10^2 = 2000\, V/m$.
Therefore,the electric field is $2000\, V/m$.

(D) The electric field components are given by the negative gradient of the potential: $E_x = -\frac{\partial V}{\partial x}$,$E_y = -\frac{\partial V}{\partial y}$,and $E_z = -\frac{\partial V}{\partial z}$.
Calculating the partial derivatives:
$E_x = -\frac{\partial}{\partial x}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(6 - 8y^2) = 8y^2 - 6$.
$E_y = -\frac{\partial}{\partial y}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(-16xy - 8 + 6z) = 16xy + 8 - 6z$.
$E_z = -\frac{\partial}{\partial z}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(6y - 8z) = 8z - 6y$.
At the origin $(x=0, y=0, z=0)$:
$E_x = 8(0)^2 - 6 = -6 \, V/m$.
$E_y = 16(0)(0) + 8 - 6(0) = 8 \, V/m$.
$E_z = 8(0) - 6(0) = 0 \, V/m$.
The magnitude of the electric field is $E = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6)^2 + 8^2 + 0^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, N/C$.
The electric force $F$ on a charge $q = 2 \, C$ is $F = qE = 2 \times 10 = 20 \, N$.

(B) The electric field components are given by the negative gradient of the potential:
$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(-kxy) = ky$
$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(-kxy) = kx$
The magnitude of the electric field intensity $E$ is given by:
$E = \sqrt{E_x^2 + E_y^2} = \sqrt{(ky)^2 + (kx)^2}$
$E = k\sqrt{x^2 + y^2}$
Since the distance from the origin is $r = \sqrt{x^2 + y^2}$,we have:
$E = kr$
Therefore,the field intensity varies directly with the distance $r$ from the origin,i.e.,$E \propto r$.

(A) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dR}$, which represents the negative of the slope of the $V-R$ graph.
For the region between $R = 4\,m$ and $R = 6\,m$, the graph is a straight line passing through $(4, 5)$ and $(6, 0)$.
The slope of this line is $m = \frac{V_2 - V_1}{R_2 - R_1} = \frac{0 - 5}{6 - 4} = \frac{-5}{2} = -2.5\,V/m$.
Therefore, the electric field at $R = 5\,m$ is $E = -(\text{slope}) = -(-2.5) = 2.5\,V/m$.

(B) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$. This means the magnitude of the electric field is equal to the slope of the $V-x$ graph.
In regions $1$,$3$,and $5$,the potential $V$ is constant,so the slope $\frac{dV}{dx} = 0$. Thus,${E_1} = {E_3} = {E_5} = 0$.
In regions $2$ and $4$,the potential changes linearly. The magnitude of the electric field is given by the slope of the lines in these regions.
From the graph,the slope of the line in region $4$ is steeper than the slope of the line in region $2$. Therefore,the magnitude of the electric field in region $4$ is greater than in region $2$,i.e.,${E_2} < {E_4}$.
Thus,the correct relation is ${E_1} = {E_3} = {E_5} = 0$ and ${E_2} < {E_4}$.

(A) The potential gradient is defined as the fall of potential per unit length along the wire.
Mathematically,it is expressed as:
Potential gradient $ = \frac{\text{Potential difference}}{\text{Length}}$
Therefore,option $A$ is correct.

(C) The relationship between force $F$ and potential energy $U$ is given by $F = - \frac{dU}{dr}$.
Given $U = - \frac{ke^2}{3r^3} = - \frac{ke^2}{3} r^{-3}$.
Now,differentiating $U$ with respect to $r$:
$F = - \frac{d}{dr} \left( - \frac{ke^2}{3} r^{-3} \right)$
$F = \frac{ke^2}{3} \frac{d}{dr} (r^{-3})$
$F = \frac{ke^2}{3} (-3 r^{-4})$
$F = - \frac{ke^2}{r^4}$.

(A) Step $1$: The electric field $\vec{E}$ is related to the potential $V$ by the gradient formula: $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Given $V = 4x^2$,we calculate the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$,$\frac{\partial V}{\partial y} = 0$,and $\frac{\partial V}{\partial z} = 0$.
Substituting these into the formula,we get $\vec{E} = -(8x) \hat{i} = -8x \hat{i} \ V/m$.
Step $2$: Evaluate the electric field at the point $(1, 0, 2)$.
At $x = 1$,$\vec{E} = -8(1) \hat{i} = -8 \hat{i} \ V/m$.
The negative sign indicates that the electric field is directed along the negative $x$-axis with a magnitude of $8 \ V/m$.

(C) The potential gradient is defined as the rate of change of potential with respect to distance,given by the formula: $E = \frac{V}{d}$.
Given:
Potential $V = 20 \ kV = 20,000 \ V = 2 \times 10^4 \ V$.
Distance $d = 2 \ mm = 2 \times 10^{-3} \ m$.
Substituting the values into the formula:
$E = \frac{2 \times 10^4 \ V}{2 \times 10^{-3} \ m} = 1 \times 10^{4 - (-3)} \ V/m = 10^7 \ V/m$.
Thus,the potential gradient is $10^7 \ V/m$.

(C) The relationship between the electric field $E$ and the electric potential $V$ is given by the negative gradient of the potential.
For the $x$-component of the electric field,the relation is $E_x = -\frac{dV}{dx}$.
This represents the rate of change of potential with respect to distance in the $x$-direction.

(B) The electric field $E_r$ is related to the potential $\phi$ by $E_r = -\frac{d\phi}{dr}$.
Given $\phi = ar^2 + b$,we have $E_r = -\frac{d}{dr}(ar^2 + b) = -2ar$.
According to Gauss's Law in differential form,the charge density $\rho$ is given by $\rho = \varepsilon_0 (\nabla \cdot E)$.
In spherical coordinates,for a radial field $E_r$,the divergence is $\nabla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 E_r)$.
Substituting $E_r = -2ar$:
$\nabla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 (-2ar)) = \frac{1}{r^2} \frac{d}{dr}(-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a$.
Therefore,the charge density is $\rho = \varepsilon_0 (-6a) = -6a\varepsilon_0$.

(A) Given: Distance $d = 5 \, mm = 5 \times 10^{-3} \, m$,Potential difference $V = 50 \, V$,Mass $m = 10^{-15} \, kg$,Charge $q = 10^{-11} \, C$.
Electric field $E = \frac{V}{d} = \frac{50}{5 \times 10^{-3}} = 10^4 \, V/m$.
The force on the particle is $F = qE$.
Using Newton's second law,$F = ma$,so $a = \frac{qE}{m}$.
Substituting the values: $a = \frac{10^{-11} \times 10^4}{10^{-15}} = \frac{10^{-7}}{10^{-15}} = 10^8 \, m/s^2$.

(C) The relationship between electric field $E$,potential difference $V$,and distance $d$ is given by $E = \frac{V}{d}$.
Given: $V = 10 \, V$ and $d = 20 \, cm = 0.2 \, m$.
Substituting the values:
$E = \frac{10}{0.2} = \frac{100}{2} = 50 \, V m^{-1}$.
Therefore,the electric field is $50 \, V m^{-1}$.

(D) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.
Given $V(x) = 20(x^2 - 4)^{-1}$.
Differentiating with respect to $x$:
$\frac{dV}{dx} = 20 \cdot (-1) \cdot (x^2 - 4)^{-2} \cdot (2x) = -\frac{40x}{(x^2 - 4)^2}$.
Therefore,$E = -\left(-\frac{40x}{(x^2 - 4)^2}\right) = \frac{40x}{(x^2 - 4)^2}$.
At $x = 4 \ \mu m$:
$E = \frac{40(4)}{(4^2 - 4)^2} = \frac{160}{(16 - 4)^2} = \frac{160}{12^2} = \frac{160}{144}$.
Simplifying the fraction by dividing by $16$:
$E = \frac{10}{9} \ V/\mu m$.
Since the value is positive,the electric field is in the $+ve \ x$-direction.

(D) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = -5$,$\frac{\partial V}{\partial y} = 3$,and $\frac{\partial V}{\partial z} = \sqrt{15}$.
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -(-5\hat{i} + 3\hat{j} + \sqrt{15}\hat{k}) = 5\hat{i} - 3\hat{j} - \sqrt{15}\hat{k}$.
The magnitude of the electric field is $|\vec{E}| = \sqrt{(5)^2 + (-3)^2 + (-\sqrt{15})^2}$.
$|\vec{E}| = \sqrt{25 + 9 + 15} = \sqrt{49} = 7$ units.

(A) The electric field $E$ between two parallel plates is given by the potential gradient: $E = \frac{|\Delta V|}{d}$.
Given,$V_1 = -10\, V$ and $V_2 = +30\, V$.
The potential difference $\Delta V = V_2 - V_1 = 30 - (-10) = 40\, V$.
The distance $d = 2\, cm = 2 \times 10^{-2}\, m$.
Substituting these values into the formula:
$E = \frac{40}{2 \times 10^{-2}} = 20 \times 10^2 = 2000\, V/m$.

(B) The electric field component $E_y$ along the $Y$-axis is given by the negative potential gradient: $E_y = -\frac{\Delta V}{\Delta y}$.
Given: $q = 4 \ \mu C = 4 \times 10^{-6} \ C$,$\Delta V = V_B - V_A = 54.8 \ V - 56 \ V = -1.2 \ V$,and $\Delta y = y_B - y_A = 12.5 \ cm - 12.3 \ cm = 0.2 \ cm = 0.2 \times 10^{-2} \ m$.
The electric field is $E_y = -\frac{-1.2}{0.2 \times 10^{-2}} = \frac{1.2}{0.002} = 600 \ V/m$.
The force component $F_y$ on the charge $q$ is $F_y = q E_y = (4 \times 10^{-6} \ C) \times (600 \ V/m) = 2400 \times 10^{-6} \ N = 24 \times 10^{-4} \ N$.

(D) The electric field $\overrightarrow{E}$ is related to the potential $V$ by $\overrightarrow{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = 6 - 8y^2$
$\frac{\partial V}{\partial y} = -16xy - 8 + 6z$
$\frac{\partial V}{\partial z} = 6y - 8z$
At the origin $(0, 0, 0)$:
$\frac{\partial V}{\partial x} = 6 - 0 = 6$
$\frac{\partial V}{\partial y} = 0 - 8 + 0 = -8$
$\frac{\partial V}{\partial z} = 0 - 0 = 0$
Thus,the electric field at the origin is $\overrightarrow{E} = -(6 \hat{i} - 8 \hat{j} + 0 \hat{k}) = -6 \hat{i} + 8 \hat{j} \, N/C$.
The magnitude of the electric field is $|\overrightarrow{E}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, N/C$.
The force on the charge $q = 2 \, C$ is $F = q|\overrightarrow{E}| = 2 \times 10 = 20 \, N$.

(A) The potential difference $V_B - V_A$ is given by the relation $V_B - V_A = - \int_{A}^{B} E \cdot dx$.
Here,$A = 10 \, m$ and $B = 20 \, m$.
$V_{20} - V_{10} = - \int_{10}^{20} \frac{100}{x^2} \, dx$.
$V_{20} - V_{10} = - 100 \left[ -\frac{1}{x} \right]_{10}^{20}$.
$V_{20} - V_{10} = 100 \left[ \frac{1}{20} - \frac{1}{10} \right]$.
$V_{20} - V_{10} = 100 \left[ \frac{1 - 2}{20} \right] = 100 \left( -\frac{1}{20} \right) = -5 \, V$.
The magnitude of the potential difference is $|V_{10} - V_{20}| = 5 \, V$.

(D) The electric field components are given by $E_x = -\frac{\partial V}{\partial x}$,$E_y = -\frac{\partial V}{\partial y}$,and $E_z = -\frac{\partial V}{\partial z}$.
Calculating the partial derivatives:
$E_x = -\frac{\partial}{\partial x}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(6 - 8y^2) = 8y^2 - 6$
$E_y = -\frac{\partial}{\partial y}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(-16xy - 8 + 6z) = 16xy + 8 - 6z$
$E_z = -\frac{\partial}{\partial z}(6x - 8xy^2 - 8y + 6yz - 4z^2) = -(6y - 8z) = 8z - 6y$
At the origin $(x=0, y=0, z=0)$:
$E_x = 8(0)^2 - 6 = -6 \ V/m$
$E_y = 16(0)(0) + 8 - 6(0) = 8 \ V/m$
$E_z = 8(0) - 6(0) = 0 \ V/m$
The magnitude of the electric field is $E = \sqrt{E_x^2 + E_y^2 + E_z^2} = \sqrt{(-6)^2 + 8^2 + 0^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ N/C$.
The electric force is $F = qE = 2 \ C \times 10 \ N/C = 20 \ N$.

(C) The relationship between electric field $E$ and potential difference $dV$ is given by $dV = -\vec{E} \cdot d\vec{r} = -E \, dr \cos \theta$.
Here,the potential increases in the direction of the electric field. The distance between the equipotential lines of $20 \, V$ and $30 \, V$ along the $x$-axis is $dr = (20 - 10) \, cm = 10 \times 10^{-2} \, m$.
The angle between the electric field vector $\vec{E}$ and the displacement vector $d\vec{r}$ along the $x$-axis is $\theta = 180^\circ - 30^\circ = 150^\circ$ (or using the perpendicular distance method,$E = \frac{dV}{dr_{\perp}}$ where $dr_{\perp} = dr \sin 30^\circ$).
Using $E = \frac{\Delta V}{\Delta r \sin \theta} = \frac{30 - 20}{0.1 \times \sin 30^\circ} = \frac{10}{0.1 \times 0.5} = \frac{10}{0.05} = 200 \, V/m$.

(A) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V$.
Given $V = axy$,the components of the electric field are:
$E_x = -\frac{\partial V}{\partial x} = -ay$
$E_y = -\frac{\partial V}{\partial y} = -ax$
The magnitude of the electric field is $E = \sqrt{E_x^2 + E_y^2}$.
Substituting the components,we get $E = \sqrt{(-ay)^2 + (-ax)^2} = \sqrt{a^2y^2 + a^2x^2} = a\sqrt{x^2 + y^2}$.
Since the distance $r$ from the origin is defined as $r = \sqrt{x^2 + y^2}$,we have $E = ar$.
Therefore,the electric field $E$ is proportional to $r$ $(E \propto r)$.

(A) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$.
Given $V = 4x^2$.
Calculating the derivative: $E = -\frac{d}{dx}(4x^2) = -8x$.
At the point $(1 \ m, 0, 2 \ m)$,the $x$-coordinate is $1 \ m$.
Substituting $x = 1$ into the expression for $E$: $E = -8(1) = -8 \ V/m$.
The negative sign indicates that the electric field is directed along the negative $X$-axis.
Therefore,the magnitude is $8 \ V/m$ along the $-X$-axis.

(A) The relationship between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.
Given $V = 5x^2 + 10x - 9$.
Taking the derivative with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) = 10x + 10$.
Therefore,$E = -(10x + 10)$.
At $x = 1 \ m$,substitute the value into the expression:
$E = -(10(1) + 10) = -(10 + 10) = -20 \ V/m$.

(B) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $V = 4x + 3y$.
The $x$-component of the electric field is $E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(4x + 3y) = -4 \, \text{V/m}$.
The $y$-component of the electric field is $E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(4x + 3y) = -3 \, \text{V/m}$.
The magnitude of the electric field is $E = \sqrt{E_x^2 + E_y^2}$.
Substituting the values,$E = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{V/m}$ (or $\text{N/C}$).

(A) The potential gradient is defined as the potential drop per unit length,given by $E = \frac{V}{L}$.
From Ohm's law,$V = iR$,where $R = \rho \frac{L}{A}$.
Substituting $R$ into the potential gradient formula:
$E = \frac{i \rho L}{A L} = \frac{i \rho}{A}$.
Given values: $i = 0.2 \, A$,$\rho = 40 \times 10^{-8} \, \Omega \, m$,and $A = 8 \times 10^{-6} \, m^2$.
Substituting these values:
$E = \frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}}$.
$E = \frac{8 \times 10^{-8}}{8 \times 10^{-6}} = 10^{-2} \, V/m$.

(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r}$.
Given $V = Q \times 10^{11} \, V$,we have $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} = Q \times 10^{11}$.
This simplifies to $\frac{1}{r} = 4 \pi \varepsilon_0 \times 10^{11}$.
The electric field $E$ at that point is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2} = \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} \right) \cdot \frac{1}{r}$.
Substituting the values,$E = (Q \times 10^{11}) \times (4 \pi \varepsilon_0 \times 10^{11})$.
Therefore,$E = 4 \pi \varepsilon_0 Q \times 10^{22} \, V/m$.

(D) The electric potential is given by $V = -x^2y - xz^3 + 4$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\vec{\nabla} V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(-x^2y - xz^3 + 4) = -2xy - z^3$
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(-x^2y - xz^3 + 4) = -x^2$
$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(-x^2y - xz^3 + 4) = -3xz^2$
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -[(-2xy - z^3)\hat{i} + (-x^2)\hat{j} + (-3xz^2)\hat{k}]$
$\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$.

(A) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V$.
Where $\nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}$.
Therefore,$\vec{E} = -\left[ \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} \right]$.
Given $V = 4x^2$,we calculate the partial derivatives:
$\frac{\partial V}{\partial x} = 8x$,$\frac{\partial V}{\partial y} = 0$,and $\frac{\partial V}{\partial z} = 0$.
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -(8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} \text{ V/m}$.
At the point $(1, 0, 2)$,we substitute $x = 1$:
$\vec{E} = -8(1) \hat{i} = -8 \hat{i} \text{ V/m}$.
The negative sign indicates that the electric field is directed along the negative $X-$ axis with a magnitude of $8 \text{ V/m}$.

(D) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Given $V(x, y, z) = 6x - 8xy - 8y + 6yz$.
Calculating partial derivatives:
$E_x = -\frac{\partial V}{\partial x} = -(6 - 8y) = -6 + 8y$.
$E_y = -\frac{\partial V}{\partial y} = -(-8x - 8 + 6z) = 8x + 8 - 6z$.
$E_z = -\frac{\partial V}{\partial z} = -(6y) = -6y$.
At point $(1, 1, 1)$:
$E_x = -6 + 8(1) = 2 \ V/m$.
$E_y = 8(1) + 8 - 6(1) = 10 \ V/m$.
$E_z = -6(1) = -6 \ V/m$.
Thus,$\vec{E} = 2\hat{i} + 10\hat{j} - 6\hat{k} \ V/m$.
The magnitude of the electric field is $|\vec{E}| = \sqrt{2^2 + 10^2 + (-6)^2} = \sqrt{4 + 100 + 36} = \sqrt{140} = 2\sqrt{35} \ V/m$.
The electric force $\vec{F} = q\vec{E}$. Given $q = 2 \ C$,the magnitude of the force is $F = q|\vec{E}| = 2 \times 2\sqrt{35} = 4\sqrt{35} \ N$.

(C) Given the potential function: $V = 6xy - y + 2yz$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation: $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(6xy - y + 2yz) = 6y$
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(6xy - y + 2yz) = 6x - 1 + 2z$
$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(6xy - y + 2yz) = 2y$
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -[ (6y)\hat{i} + (6x - 1 + 2z)\hat{j} + (2y)\hat{k} ]$.
Now,evaluating at the point $(1, 1, 0)$:
$\vec{E}_{(1, 1, 0)} = -[ (6 \times 1)\hat{i} + (6 \times 1 - 1 + 2 \times 0)\hat{j} + (2 \times 1)\hat{k} ]$
$\vec{E}_{(1, 1, 0)} = -(6\hat{i} + 5\hat{j} + 2\hat{k}) \ N/C$.

(A) The electric field $\vec{E}$ is uniform and directed along the positive $x$-axis.
The relationship between electric field and electric potential is given by $\vec{E} = -\nabla V$,which implies that the electric potential $V$ decreases in the direction of the electric field.
Since the electric field is directed towards the positive $x$-axis,the potential decreases as we move from left to right along the $x$-axis.
Comparing the $x$-coordinates of the given points:
Point $C$ is at $x = -a$
Points $B$ and $D$ are at $x = 0$
Point $A$ is at $x = a$
Since the potential decreases as $x$ increases,the potential will be maximum at $C$ and minimum at $A$.

(B) Given,$V = k[2x^2 - y^2 + z^2]$.
The electric field is given by the negative gradient of the potential: $\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}\right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = k(4x)$
$\frac{\partial V}{\partial y} = k(-2y)$
$\frac{\partial V}{\partial z} = k(2z)$
Thus,$\vec{E} = -k(4x \hat{i} - 2y \hat{j} + 2z \hat{k})$.
At the point $(1, 1, 1)$,the electric field is:
$\vec{E}_{(1,1,1)} = -k(4(1) \hat{i} - 2(1) \hat{j} + 2(1) \hat{k}) = -k(4 \hat{i} - 2 \hat{j} + 2 \hat{k})$.
The magnitude of the electric field is:
$|\vec{E}| = \sqrt{(-4k)^2 + (2k)^2 + (-2k)^2} = \sqrt{16k^2 + 4k^2 + 4k^2} = \sqrt{24k^2} = 2k\sqrt{6}$.

(D) The relationship between electric field $E_x$ and potential $V$ is given by $E_x = -\frac{dV}{dx}$.
This means the electric field is the negative of the slope of the $V-x$ graph.
At point $A$, the slope $\frac{dV}{dx}$ is positive, so $E_x = -(\text{positive}) = \text{negative}$. Thus, the electric field at $A$ is along the negative $x$-axis.
At point $B$, the graph reaches its peak, so the slope $\frac{dV}{dx} = 0$, which means $E_x = 0$.
At point $C$, the slope $\frac{dV}{dx}$ is negative, so $E_x = -(\text{negative}) = \text{positive}$. Thus, the electric field at $C$ is along the positive $x$-axis.
Therefore, the correct statement is that the $x$-component at point $C$ is along the positive $x$-axis.

(A) Given the potential function $V(x) = \frac{20}{x^2 - 4} \text{ volt}$.
The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$.
$E = -\frac{d}{dx} \left( 20(x^2 - 4)^{-1} \right) = -20 \cdot (-1)(x^2 - 4)^{-2} \cdot (2x) = \frac{40x}{(x^2 - 4)^2}$.
Now,substituting $x = 4 \mu m$ into the expression for $E$:
$E = \frac{40(4)}{(4^2 - 4)^2} = \frac{160}{(16 - 4)^2} = \frac{160}{12^2} = \frac{160}{144}$.
Simplifying the fraction by dividing both numerator and denominator by $16$,we get $E = \frac{10}{9} \text{ V/}\mu m$.
Since the result is positive,the electric field is in the $+ve \ x$ direction.

(B) The potential difference between two points in an electric field is given by the relation $dV = -\vec{E} \cdot d\vec{x}$.
Integrating this expression from the origin $(x = 0)$ to the point $x = 2 \ m$:
$V_A - V_O = -\int_{0}^{2} 30x^2 dx$
$V_A - V_O = -[10x^3]_{0}^{2}$
$V_A - V_O = -(10(2)^3 - 10(0)^3)$
$V_A - V_O = -(10 \times 8) = -80 \ V$.

(B) The relationship between the electric field $(E)$ and the potential $(V)$ is given by the formula $E_x = -\frac{dV}{dx}$.
This means the magnitude of the $x$-component of the electric field,$|E_x|$,is equal to the absolute value of the slope of the $V-X$ graph,i.e.,$|E_x| = |\frac{dV}{dx}|$.
To find the region where the magnitude of the electric field is maximum,we need to identify the region where the slope of the $V-X$ graph is steepest (i.e.,has the largest absolute value).
Looking at the graph:
- In regions $1$ and $4$,the potential is constant,so the slope is $0$,which means $E_x = 0$.
- In regions $2$ and $3$,the potential changes with distance,so the slope is non-zero.
- Comparing the steepness of the lines in regions $2$ and $3$,the line in region $2$ is much steeper than the line in region $3$.
Therefore,the magnitude of the slope is maximum in region $2$,which implies the magnitude of the electric field is maximum in region $2$.

(D) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Given $V = -5x + 3y + \sqrt{15}z$.
Calculating the partial derivatives:
$E_x = -\frac{\partial V}{\partial x} = -(-5) = 5$
$E_y = -\frac{\partial V}{\partial y} = -(3) = -3$
$E_z = -\frac{\partial V}{\partial z} = -(\sqrt{15}) = -\sqrt{15}$
Thus,$\vec{E} = 5\hat{i} - 3\hat{j} - \sqrt{15}\hat{k}$.
The magnitude of the electric field is $|\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2}$.
$|\vec{E}| = \sqrt{(5)^2 + (-3)^2 + (-\sqrt{15})^2} = \sqrt{25 + 9 + 15} = \sqrt{49} = 7$.

(B) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$.
This implies that the magnitude of the electric field is equal to the magnitude of the potential gradient,$|E| = |\frac{dV}{dx}|$.
In the given diagram,the electric field lines are closer together near $X$ and become more spread out (uniform) as we move towards $Y$.
This indicates that the electric field strength $|E|$ is higher near $X$ and decreases as we move towards $Y$.
Since $|E| = |\frac{dV}{dx}|$,the slope of the potential $V$ versus distance $d$ graph must be steeper near $X$ and become less steep as we move towards $Y$.
Among the given options,the graph in option $B$ shows a steep decrease in $V$ near $X$ (high slope) and a shallower decrease as we move towards $Y$ (lower slope),which correctly represents the variation of potential.

(A) The electric field $E_x$ is related to the potential $V$ by the relation $E_x = -\frac{dV}{dx}$.
$1$. For $x < -d$,$V = 0$,so $E_x = -\frac{d(0)}{dx} = 0$.
$2$. For $-d \le x < 0$,$V = -V_0(1 + \frac{x}{d})^2$.
$E_x = -\frac{d}{dx}[-V_0(1 + \frac{x}{d})^2] = V_0 \cdot 2(1 + \frac{x}{d}) \cdot \frac{1}{d} = \frac{2V_0}{d}(1 + \frac{x}{d})$.
At $x = -d$,$E_x = 0$. At $x = 0$,$E_x = \frac{2V_0}{d}$.
$3$. For $0 \le x < d$,$V = -V_0(1 + 2\frac{x}{d})$.
$E_x = -\frac{d}{dx}[-V_0(1 + 2\frac{x}{d})] = V_0 \cdot \frac{2}{d} = \frac{2V_0}{d}$.
This is a constant value.
$4$. For $x \ge d$,$V = -3V_0$,so $E_x = -\frac{d(-3V_0)}{dx} = 0$.
Comparing these results with the given options,the graph in option $A$ correctly represents $E_x$ as a function of $x$.

(B) The equation of the equipotential lines is $V = ax + by + c$. From the graph,for $V = 2 \, V$,the line passes through $(6, 2)$ and $(4, 1)$. The slope is $m = \frac{1-2}{4-6} = \frac{-1}{-2} = 0.5$. The equation is $y - 1 = 0.5(x - 4) \Rightarrow y = 0.5x - 1 \Rightarrow 0.5x - y = 1$. Multiplying by $2$,we get $x - 2y = 2$. Thus,$V = k(x - 2y)$. For $V = 2$,$2 = k(6 - 2(2)) = 2k \Rightarrow k = 1$. So,$V = x - 2y$.
The electric field components are $E_x = -\frac{\partial V}{\partial x} = -1 \, V/\text{cm} = -100 \, V/m$ and $E_y = -\frac{\partial V}{\partial y} = -(-2) = +2 \, V/\text{cm} = +200 \, V/m$.
Thus,$E_x = -100 \, V/m$ and $E_y = +200 \, V/m$.

(D) The electric field vector is given by $\vec{E} = E \cos 45^\circ \hat{i} + E \sin 45^\circ \hat{j} = \frac{400}{\sqrt{2}} (\hat{i} + \hat{j}) \, V/m$.
The potential difference between two points $A(0, 0.02 \, m)$ and $B(0.03 \, m, 0)$ is given by $V_A - V_B = \int_{B}^{A} \vec{E} \cdot d\vec{r}$.
Alternatively,using $V_A - V_B = \vec{E} \cdot \vec{r}_{BA}$,where $\vec{r}_{BA} = \vec{r}_A - \vec{r}_B = (0 - 0.03) \hat{i} + (0.02 - 0) \hat{j} = -0.03 \hat{i} + 0.02 \hat{j} \, m$.
$V_A - V_B = \left[ \frac{400}{\sqrt{2}} (\hat{i} + \hat{j}) \right] \cdot (-0.03 \hat{i} + 0.02 \hat{j})$
$V_A - V_B = \frac{400}{\sqrt{2}} (-0.03 + 0.02) = \frac{400}{\sqrt{2}} (-0.01) = -\frac{4}{\sqrt{2}} = -2\sqrt{2} \approx -2.828 \, V$.
Given the options,the magnitude or the closest value is $2.8 \, V$ (considering the path direction or potential drop definition).

(A) The electric field $\overrightarrow{E}$ is related to the electric potential $V$ by the gradient formula: $\overrightarrow{E} = -\nabla V = -\left[ \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} \right]$.
Given $V = 4x^2$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$
$\frac{\partial V}{\partial y} = 0$
$\frac{\partial V}{\partial z} = 0$
Substituting these into the electric field formula:
$\overrightarrow{E} = -[\hat{i}(8x) + \hat{j}(0) + \hat{k}(0)] = -8x \hat{i}$.
At the point $(1 \, m, 0, 2 \, m)$,the value of $x$ is $1 \, m$.
Therefore,$\overrightarrow{E} = -8(1) \hat{i} = -8 \hat{i} \, V/m$.
The negative sign indicates that the electric field is directed along the negative $x$-axis with a magnitude of $8 \, V/m$.

(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.
Given $V = b - ar^2$,we have $E = -\frac{d}{dr}(b - ar^2) = 2ar$.
According to Gauss's Law in differential form,the charge density $\rho$ is given by $\rho = \varepsilon_0 (\nabla \cdot E)$.
In spherical coordinates,for a radial field $E(r)$,the divergence is $\nabla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 E)$.
Substituting $E = 2ar$,we get $\nabla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot 2ar) = \frac{1}{r^2} \frac{d}{dr}(2ar^3) = \frac{1}{r^2} (6ar^2) = 6a$.
Therefore,the charge density is $\rho = \varepsilon_0 (6a) = 6a\varepsilon_0$.

(B) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $V = 5x^2 + 5xy$.
Calculating the partial derivatives:
$E_x = -\frac{\partial V}{\partial x} = -(10x + 5y)$.
At point $(1, -2)$,$E_x = -(10(1) + 5(-2)) = -(10 - 10) = 0$.
$E_y = -\frac{\partial V}{\partial y} = -(5x) = -5x$.
At point $(1, -2)$,$E_y = -5(1) = -5$.
Therefore,the electric field at $(1, -2)$ is $\vec{E} = 0 \hat{i} - 5 \hat{j} = -5 \hat{j} \, V/m$.