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

(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$. The magnitude of the electric field is $|E| = |\frac{dV}{dx}|$,which represents the slope of the $V-x$ graph.
In region $A$ ($x=0$ to $1$ m),$V$ is constant ($2$ $V$),so the slope $\frac{dV}{dx} = 0$. Thus,$E_{A} = 0$.
In region $B$ ($x=1$ to $2$ m),the slope is $\frac{4-2}{2-1} = 2$ $V$/m. Thus,$|E_{B}| = 2$ $V$/m.
In region $C$ ($x=2$ to $4$ m),$V$ is constant ($4$ $V$),so the slope $\frac{dV}{dx} = 0$. Thus,$E_{C} = 0$.
In region $D$ ($x=4$ to $5$ m),the slope is $\frac{0-4}{5-4} = -4$ $V$/m. Thus,$|E_{D}| = |-4| = 4$ $V$/m.
Comparing the magnitudes: $E_{A} = E_{C} = 0$ and $E_{B} = 2$ $V$/m,$E_{D} = 4$ $V$/m. Therefore,$E_{A} = E_{C}$ and $E_{B} < E_{D}$.

(D) The electric potential is given by $V = 3x^2$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3x^2) = 6x$.
$\frac{\partial V}{\partial y} = 0$.
$\frac{\partial V}{\partial z} = 0$.
Thus,$\vec{E} = -6x \hat{i}$.
At the point $(2, 0, 1)$,the $x$-coordinate is $2$.
Substituting $x = 2$ into the expression for $\vec{E}$:
$\vec{E} = -6(2) \hat{i} = -12 \hat{i} \ Vm^{-1}$.
The magnitude of the electric field is $-12 \ Vm^{-1}$ along the $x$-direction.

(A) The electric field intensity $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V$.
Since the potential $V$ depends only on $x$,the electric field is given by $E_x = -\frac{dV}{dx}$.
Given $V = 3x^2 + 5$.
Calculating the derivative: $\frac{dV}{dx} = \frac{d}{dx}(3x^2 + 5) = 6x$.
Therefore,$E_x = -6x$.
At the point $(-2, 1, 0)$,the $x$-coordinate is $-2$.
Substituting $x = -2$ into the expression for $E_x$:
$E_x = -6(-2) = +12 \ Vm^{-1}$.
Thus,the intensity of the electric field at $(-2, 1, 0)$ is $+12 \ Vm^{-1}$.

(C) In a uniform electric field, the electric potential decreases in the direction of the electric field lines.
Let the electric field be directed along the positive $x$-axis.
The potential at any point $(x, y)$ is given by $V = -E \cdot x + \text{constant}$.
Comparing the $x$-coordinates of the points $A, B$, and $C$ from the figure:
Point $A$ is the furthest to the right, so it has the largest $x$-coordinate.
Points $B$ and $C$ have the same $x$-coordinate, so $V_B = V_C$.
Since $A$ is further along the direction of the electric field than $B$ and $C$, the potential at $A$ is the lowest.
Therefore, $V_B = V_C > V_A$.
Looking at the options provided, the relationship $V_C > V_A > V_B$ is not strictly correct as $V_B = V_C$. However, based on the relative positions, $C$ and $B$ are at the same potential, and $A$ is at a lower potential. If we assume the points are slightly offset in the $x$-direction such that $C$ is behind $B$, then $V_C > V_B > V_A$. Given the standard interpretation of such diagrams, $C$ and $B$ are at the same potential, and $A$ is at the lowest. Among the choices, $V_C > V_B > V_A$ is the most logical representation of the potential gradient.

(B) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V$.
Given $V = 20|\vec{r}| = 20\sqrt{x^2 + y^2 + z^2}$.
Calculating the gradient $\nabla V = \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}$.
$\frac{\partial V}{\partial x} = 20 \cdot \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \cdot 2x = \frac{20x}{|\vec{r}|}$.
Similarly,$\frac{\partial V}{\partial y} = \frac{20y}{|\vec{r}|}$ and $\frac{\partial V}{\partial z} = \frac{20z}{|\vec{r}|}$.
Thus,$\vec{E} = -\frac{20}{|\vec{r}|} (x \hat{i} + y \hat{j} + z \hat{k}) = -\frac{20}{|\vec{r}|} \vec{r} = -20 \hat{r}$.
At the point $(4, 3, -5)$,the magnitude $|\vec{r}| = \sqrt{4^2 + 3^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.
Substituting the values: $\vec{E} = -\frac{20}{5\sqrt{2}} (4 \hat{i} + 3 \hat{j} - 5 \hat{k}) = -\frac{4}{\sqrt{2}} (4 \hat{i} + 3 \hat{j} - 5 \hat{k}) = -2\sqrt{2} (4 \hat{i} + 3 \hat{j} - 5 \hat{k})$.
This simplifies to $-\sqrt{2} (8 \hat{i} + 6 \hat{j} - 10 \hat{k})$.
Therefore,the correct option is $B$.

(D) Given: $\vec{E} = (30 \hat{i} + 40 \hat{j}) \text{ NC}^{-1}$ and $V(0,0) = 0 \text{ V}$.
We know that the relation between electric field and electric potential is given by $dV = -\vec{E} \cdot d\vec{r}$.
Here,$d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$.
Integrating from the origin $(0,0)$ to the point $(1,2)$:
$\int_{V(0,0)}^{V(1,2)} dV = -\int_{(0,0)}^{(1,2)} (30 \hat{i} + 40 \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$
$V(1,2) - V(0,0) = -\left[ \int_{0}^{1} 30 dx + \int_{0}^{2} 40 dy \right]$
$V(1,2) - 0 = -[30(1) + 40(2)]$
$V(1,2) = -(30 + 80) = -110 \text{ V}$.

(C) The electric field $\vec{E}$ is given by the negative gradient of the potential $V$: $\vec{E} = -\vec{\nabla} V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $V = \frac{1}{2} y^2 - 2x$.
Calculating partial derivatives:
$\frac{\partial V}{\partial x} = -2$
$\frac{\partial V}{\partial y} = y$
Substituting these into the electric field formula:
$\vec{E} = -(-2 \hat{i} + y \hat{j}) = 2 \hat{i} - y \hat{j}$.
At the point $(x = 1 \text{ m}, y = 1 \text{ m})$:
$\vec{E} = 2 \hat{i} - (1) \hat{j} = 2 \hat{i} - \hat{j} \text{ Vm}^{-1}$.

(C) The electric field is given by $E = 5x \hat{i}$.
The potential difference between two points is given by $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.
Since the electric field is only in the $X$-direction, the $Y$-axis (where $x=0$) is an equipotential line. Thus, the potential at point $A(0, 5)$ is the same as the potential at the origin $C(0, 0)$.
$V_A = V_C$.
Now, we calculate the potential difference between $C(0, 0)$ and $B(2, 0)$:
$V_B - V_C = -\int_{0}^{2} E_x dx = -\int_{0}^{2} 5x dx$.
$V_B - V_C = -5 \left[ \frac{x^2}{2} \right]_{0}^{2} = -5 \left( \frac{4}{2} - 0 \right) = -5(2) = -10 \text{ V}$.
Since $V_A = V_C$, we have $V_B - V_A = -10 \text{ V}$.

(A) The electric field $\vec{E}$ is related to the electric potential $\phi$ by the relation $\vec{E} = -\nabla \phi$.
Since the potential $\phi$ depends only on $x$,the electric field is given by $\vec{E} = -\frac{\partial \phi}{\partial x} \hat{i}$.
Given $\phi(x) = \phi_0 \frac{x_0}{x} = (8 \ V)(5 \ m) \frac{1}{x} = \frac{40}{x} \ V \cdot m$.
Differentiating with respect to $x$: $\frac{\partial \phi}{\partial x} = 40 \frac{d}{dx}(x^{-1}) = 40 (-x^{-2}) = -\frac{40}{x^2}$.
Therefore,$\vec{E} = -(-\frac{40}{x^2}) \hat{i} = \frac{40}{x^2} \hat{i} \ V/m$.
At the point $(10 \ m, 5 \ m, 5 \ m)$,the $x$-coordinate is $10 \ m$.
Substituting $x = 10 \ m$ into the expression for $\vec{E}$: $\vec{E} = \frac{40}{10^2} \hat{i} = \frac{40}{100} \hat{i} = 0.40 \ Vm^{-1} \hat{i}$.

(C) The relation between electric field $\vec{E}$ and potential $V$ is given by $\vec{E} = -\nabla V$,which implies $dV = -\vec{E} \cdot d\vec{r}$.
Given $\vec{E} = A(x \hat{i} + y \hat{j})$,we have $dV = -A(x dx + y dy)$.
Integrating both sides,we get $V = -A \int x dx - A \int y dy = -A \frac{x^2}{2} - A \frac{y^2}{2} + C$,where $C$ is the integration constant.
Substituting $A = 10 \ Vm^{-2}$,we get $V = -5(x^2 + y^2) + C$.
Given that the potential at $(10 \ m, 20 \ m)$ is zero,we have $0 = -5(10^2 + 20^2) + C$.
$0 = -5(100 + 400) + C \Rightarrow 0 = -5(500) + C \Rightarrow C = 2500 \ V$.
Thus,the potential function is $V(x, y) = -5(x^2 + y^2) + 2500$.
At the origin $(0, 0)$,the potential is $V(0, 0) = -5(0^2 + 0^2) + 2500 = 2500 \ V$.

(D) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.
Given $\sigma = 7 \times 10^{-7} \text{ C m}^{-2}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}$.
We know that $\varepsilon_0 = \frac{1}{4 \pi \times 9 \times 10^9} \approx 8.85 \times 10^{-12} \text{ F m}^{-1}$.
The relation between electric field and potential difference $\Delta V$ for a distance $\Delta r$ is $|E| = \frac{\Delta V}{\Delta r}$.
Substituting the values:
$\frac{\sigma}{2 \varepsilon_0} = \frac{\Delta V}{\Delta r}$
$\Delta r = \frac{\Delta V \times 2 \varepsilon_0}{\sigma} = \frac{19.8 \times 2 \times 8.85 \times 10^{-12}}{7 \times 10^{-7}}$
$\Delta r = \frac{19.8 \times 17.7 \times 10^{-12}}{7 \times 10^{-7}} \approx 5 \times 10^{-4} \text{ m} = 0.5 \text{ mm}$.

(C) The electric field $\vec{E}$ is related to the potential gradient by the relation $\vec{E} = -\nabla V$.
The magnitude of the potential gradient is maximum in the direction of the electric field.
By definition,an equipotential surface is a surface where the electric potential $V$ is constant at every point.
For any displacement $d\vec{r}$ along the equipotential surface,the change in potential $dV = -\vec{E} \cdot d\vec{r} = 0$.
This implies that the electric field vector $\vec{E}$ must be perpendicular to the surface at every point.
Since the maximum potential gradient points in the direction of the electric field,the angle between the maximum potential gradient and the equipotential surface is $90^{\circ}$ or $\frac{\pi}{2}$ radians.

(A) The relationship between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.
From the given graph,it is observed that in the region between $x = 1 \ m$ and $x = 3 \ m$,the electric potential $V$ is constant $(V = 2 \ V)$.
Since the potential is constant in this region,the rate of change of potential with respect to distance is zero,i.e.,$\frac{dV}{dx} = 0$.
Therefore,the electric field at $x = 2 \ m$ is $E = -\frac{dV}{dx} = 0 \ V/m$.

(B) The potential difference $\Delta V = V_B - V_A$ is given by the line integral: $\Delta V = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.
Given $\vec{E} = x \hat{i} - 2y \hat{j} + z \hat{k}$ and $d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k}$.
$\Delta V = -\int_{(2,1,0)}^{(0,2,4)} (x \hat{i} - 2y \hat{j} + z \hat{k}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k})$
$\Delta V = -[\int_{2}^{0} x \ dx - \int_{1}^{2} 2y \ dy + \int_{0}^{4} z \ dz]$
Evaluating the integrals:
$\int_{2}^{0} x \ dx = [\frac{x^2}{2}]_{2}^{0} = 0 - 2 = -2$
$\int_{1}^{2} 2y \ dy = [y^2]_{1}^{2} = 4 - 1 = 3$
$\int_{0}^{4} z \ dz = [\frac{z^2}{2}]_{0}^{4} = 8 - 0 = 8$
Substituting these values:
$\Delta V = -[-2 - 3 + 8] = -[3] = -3 \ V$.
The magnitude of the potential difference is $|\Delta V| = 3 \ V$.

(C) The relation between electric field $E$ and electric potential $V$ is given by $E = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $E = 3 \hat{i} + 4y \hat{j}$,we have:
$-\frac{\partial V}{\partial x} = 3 \implies \frac{\partial V}{\partial x} = -3$
$-\frac{\partial V}{\partial y} = 4y \implies \frac{\partial V}{\partial y} = -4y$
Integrating these partial derivatives:
$V(x, y) = \int -3 \ dx = -3x + f(y)$
$V(x, y) = \int -4y \ dy = -2y^2 + g(x)$
Combining these,the general potential function is $V(x, y) = -(3x + 2y^2) + C$.
Given that the potential at the origin $(0, 0)$ is $0$,we have $V(0, 0) = -(3(0) + 2(0)^2) + C = 0$,which implies $C = 0$.
Thus,$V(x, y) = -(3x + 2y^2)$.
At point $(2, 1)$,the potential is $V(2, 1) = -(3(2) + 2(1)^2) = -(6 + 2) = -8 \ V$.

(D) The electric field $E$ between the parallel plates is given by $E = \frac{V}{d}$.
Given, $V = 10^4 \,V$ and $d = 0.5 \,cm = 0.5 \times 10^{-2} \,m$.
Therefore, $E = \frac{10^4}{0.5 \times 10^{-2}} = 2 \times 10^6 \,V/m$.
The force $F$ on an electron of charge $e = 1.6 \times 10^{-19} \,C$ is given by $F = eE$.
$F = (1.6 \times 10^{-19} \,C) \times (2 \times 10^6 \,V/m) = 3.2 \times 10^{-13} \,N$.

(C) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right)$.
Given $V = \frac{1}{2}y^2 - 2x$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (\frac{1}{2}y^2 - 2x) = -2$.
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (\frac{1}{2}y^2 - 2x) = y$.
Thus,$\vec{E} = -(-2 \hat{i} + y \hat{j}) = 2 \hat{i} - y \hat{j}$.
At the point $(x = 1 \text{ m}, y = 1 \text{ m})$:
$\vec{E} = 2 \hat{i} - (1) \hat{j} = 2 \hat{i} - \hat{j} \text{ V m}^{-1}$.

(C) Given: Electric field $E = 500 \ NC^{-1}$,Electric potential $V = 30 \ V$.
The relationship between electric field and potential for a point charge is $E = \frac{V}{r}$.
Therefore,the distance $r$ is $r = \frac{V}{E} = \frac{30}{500} = 0.06 \ m$.
The formula for electric potential is $V = \frac{kq}{r}$,where $k = 9 \times 10^9 \ Nm^2C^{-2}$.
Rearranging for charge $q$: $q = \frac{V \cdot r}{k}$.
Substituting the values: $q = \frac{30 \times 0.06}{9 \times 10^9} = \frac{1.8}{9 \times 10^9} = 0.2 \times 10^{-9} \ C$.
Thus,$q = 2 \times 10^{-10} \ C$.

(B) The relation between the electric field $\vec{E}$ and the electric potential $\phi$ is given by $\vec{E} = -\nabla \phi$.
Thus,$d\phi = -\vec{E} \cdot d\vec{r}$.
Given $\vec{E} = a(y \hat{i} + x \hat{j})$ and $d\vec{r} = dx \hat{i} + dy \hat{j}$.
Substituting these into the expression:
$d\phi = -a(y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j})$
$d\phi = -a(y dx + x dy)$
We know that $d(xy) = y dx + x dy$.
Therefore,$d\phi = -a d(xy)$.
Integrating both sides,we get:
$\phi = -a \int d(xy) = -axy + C$.
Comparing this with the given options,option $B$ is the correct answer.

(B) The electrostatic potential $V$ inside a charged sphere is given by $V = A r^2 + B$.
The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.
Substituting the given expression for $V$:
$E = -\frac{d}{dr}(A r^2 + B) = -2 A r$.
According to Gauss's Law in differential form,the volume charge density $\rho$ is related to the electric field $E$ by the equation $\nabla \cdot E = \frac{\rho}{\varepsilon_0}$.
For a spherically symmetric distribution,this becomes $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{\rho}{\varepsilon_0}$.
Substituting $E = -2 A r$:
$\frac{\rho}{\varepsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot (-2 A r)) = \frac{1}{r^2} \frac{d}{dr}(-2 A r^3) = \frac{1}{r^2} (-6 A r^2) = -6 A$.
Therefore,the charge density is $\rho = -6 A \varepsilon_0$.

(B) The electric field component $E_{x}$ is given by the negative gradient of the potential with respect to $x$,i.e.,$E_{x} = -\frac{dV}{dx}$.
From the graph,the potential difference $\Delta V = V_{2} - V_{1} = 4 \ V - 2 \ V = 2 \ V$.
The distance along the $x$-axis between these two equipotential lines is $\Delta x = 4 \ cm - 2 \ cm = 2 \ cm = 0.02 \ m$.
Therefore,the magnitude of the $x$-component of the electric field is $E_{x} = -\frac{\Delta V}{\Delta x} = -\frac{2 \ V}{0.02 \ m} = -100 \ V/m$.
Thus,the correct option is $B$.

(A) The relation between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.
This means the electric field is the negative of the slope of the $V-X$ graph.
$1$. For the interval $X \in [-4, -2]$: The slope is $\frac{10 - 0}{-2 - (-4)} = \frac{10}{2} = 5$. Thus,$E = -5$.
$2$. For the interval $X \in [-2, 2]$: The potential is constant $(V = 10)$,so the slope is $0$. Thus,$E = 0$.
$3$. For the interval $X \in [2, 7]$: The slope is $\frac{0 - 10}{7 - 2} = \frac{-10}{5} = -2$. Thus,$E = -(-2) = 2$.
Comparing these values with the given options,option $A$ represents $E = -5$ for $X \in [-4, -2]$,$E = 0$ for $X \in [-2, 2]$,and $E = 2$ for $X \in [2, 7]$.

(A) The electrostatic potential is given by $V = ar^3 + b$.
The electric field $E$ is related to the potential by $E = -\frac{dV}{dr}$.
Differentiating $V$ with respect to $r$,we get $E = -\frac{d}{dr}(ar^3 + b) = -3ar^2$.
According to Gauss's Law,the total charge $q_{enc}$ enclosed in a sphere of radius $r=1$ is given by $q_{enc} = \epsilon_0 \oint E \cdot dA$.
For a spherical surface of radius $r=1$,the area $A = 4\pi r^2 = 4\pi(1)^2 = 4\pi$.
Substituting the values at $r=1$,$E = -3a(1)^2 = -3a$.
Thus,$q_{enc} = \epsilon_0 \times (-3a) \times 4\pi = -12\pi a \epsilon_0$.
Comparing this with the given expression $\alpha \times \pi a \epsilon_0$,we find $\alpha = -12$.

(C) The electric field is given by $\vec{E} = 10x\hat{i} + 5y\hat{j}$.
We know that the relation between electric field and potential is $\Delta V = -\int \vec{E} \cdot d\vec{r}$.
Let $V_0$ be the potential at the origin $(0, 0)$ and $V_P$ be the potential at $(10, 20)$.
$V_P - V_0 = -\int_{(0,0)}^{(10,20)} (10x\hat{i} + 5y\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$.
$500 - V_0 = -\int_{0}^{10} 10x \ dx - \int_{0}^{20} 5y \ dy$.
$500 - V_0 = -[5x^2]_0^{10} - [\frac{5y^2}{2}]_0^{20}$.
$500 - V_0 = -[5(100) - 0] - [\frac{5(400)}{2} - 0]$.
$500 - V_0 = -500 - 1000$.
$500 - V_0 = -1500$.
$V_0 = 500 + 1500 = 2000 \ V$.

(A) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j})$.
Given $V = 5x^2 - 5y^2$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(5x^2 - 5y^2) = 10x$.
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(5x^2 - 5y^2) = -10y$.
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -(10x\hat{i} - 10y\hat{j}) = -10x\hat{i} + 10y\hat{j}$.
At the point $(2, 3) \text{ m}$,substituting $x = 2$ and $y = 3$:
$\vec{E} = -10(2)\hat{i} + 10(3)\hat{j} = -20\hat{i} + 30\hat{j} \text{ V/m}$.