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

(B) The potential difference $V_{AB} = V_B - V_A$ is given by the line integral of the electric field: $V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.
Given $\vec{E} = \hat{i} + 2\hat{j} + \hat{k}$ and the displacement vector $d\vec{r} = dx\hat{i} + dy\hat{j} + dz\hat{k}$.
The points are $A(0, 0, 0)$ and $B(2, 3, 4)$.
$V_A - V_B = \int_{A}^{B} \vec{E} \cdot d\vec{r} = \int_{(0,0,0)}^{(2,3,4)} (\hat{i} + 2\hat{j} + \hat{k}) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k})$.
$V_A - V_B = \int_{0}^{2} dx + \int_{0}^{3} 2dy + \int_{0}^{4} dz$.
$V_A - V_B = [x]_0^2 + 2[y]_0^3 + [z]_0^4 = 2 + 2(3) + 4 = 2 + 6 + 4 = 12 \text{ V}$.
Since the question asks for the potential difference between $A$ and $B$,we consider the magnitude or the path integral $V_A - V_B = 12 \text{ V}$.

(A) The relationship between electric field $\overrightarrow{E}$ and electric potential $V$ is given by the formula: $dV = -\overrightarrow{E} \cdot d\overrightarrow{r}$.
For a one-dimensional field along the $x$-axis,this becomes: $dV = -E_x dx$.
To find the potential difference $V_A - V_O$,we integrate from the origin $(x = 0)$ to point $A$ $(x = 2 \, m)$:
$V_A - V_O = -\int_{0}^{2} E_x dx$.
Substituting the given electric field $E_x = 30x^2$:
$V_A - V_O = -\int_{0}^{2} 30x^2 dx$.
Evaluating the integral:
$V_A - V_O = -30 \left[ \frac{x^3}{3} \right]_{0}^{2}$.
$V_A - V_O = -30 \left( \frac{2^3}{3} - \frac{0^3}{3} \right)$.
$V_A - V_O = -30 \left( \frac{8}{3} \right) = -10 \times 8 = -80 \, V$.

(B) The electric field is given as $\vec{E} = 20\hat{i}\, N/C$.
For a uniform electric field,the potential difference between two points $A$ and $B$ is given by the relation $\Delta V = V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.
Here,the displacement vector $\vec{r}_{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} = (6 - 4)\hat{i} + (5 - 2)\hat{j} = 2\hat{i} + 3\hat{j}$.
Substituting the values,we get $V_B - V_A = -(20\hat{i}) \cdot (2\hat{i} + 3\hat{j})$.
$V_B - V_A = -20 \times 2 = -40\, V$.

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

(C) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$,which is the negative of the slope of the $V-x$ graph.
At $x = 13\,m$,the point lies on the line segment between $x = 12\,m$ and $x = 16\,m$.
The coordinates of the endpoints of this segment are $(12, 30)$ and $(16, 10)$.
The slope of this line segment is $\text{slope} = \frac{V_2 - V_1}{x_2 - x_1} = \frac{10 - 30}{16 - 12} = \frac{-20}{4} = -5\,V/m$.
Therefore,the electric field $E = -(\text{slope}) = -(-5) = 5\,V/m$.

(A) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.
Given $V = 5x^2 + 10x - 9$.
Differentiating $V$ with respect to $x$,we get $\frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) = 10x + 10$.
Substituting this into the electric field formula: $E = -(10x + 10)$.
At $x = 1 \, m$,the electric field is $E = -(10(1) + 10) = -(10 + 10) = -20 \, V m^{-1}$.

(B) Using Poisson's equation for a potential that depends only on $z$: $\frac{d^2V}{dz^2} = -\frac{\rho}{\varepsilon_0}$.
For $|z| < 1 \ m$,$V(z) = 30 - 5z^2$. Therefore,$\frac{dV}{dz} = -10z$ and $\frac{d^2V}{dz^2} = -10$.
Substituting this into the equation: $-10 = -\frac{\rho_0}{\varepsilon_0} \implies \rho_0 = 10 \varepsilon_0$.
For $|z| > 1 \ m$,$V(z) = 35 - 10|z|$. For $z > 1$,$V(z) = 35 - 10z$,so $\frac{dV}{dz} = -10$ and $\frac{d^2V}{dz^2} = 0$. Thus,$\rho = 0$.
For $z < -1$,$V(z) = 35 + 10z$,so $\frac{dV}{dz} = 10$ and $\frac{d^2V}{dz^2} = 0$. Thus,$\rho = 0$.
Therefore,$\rho_0 = 10 \varepsilon_0$ for $|z| \le 1 \ m$ and $\rho_0 = 0$ elsewhere.

(A) The relation between electric field $\vec{E}$ and electric potential $V$ is given by $\vec{E} = -\nabla V$.
Thus,$dV = -\vec{E} \cdot d\vec{r} = -(E_x dx + E_y dy)$.
Given $\vec{E} = (25 \hat{i} + 30 \hat{j}) \, NC^{-1}$,we have $E_x = 25$ and $E_y = 30$.
To find the potential at $(2, 2)$ relative to the origin $(0, 0)$,we integrate:
$V(2, 2) - V(0, 0) = -\int_{(0,0)}^{(2,2)} (25 dx + 30 dy)$.
Since $V(0, 0) = 0$,we have:
$V(2, 2) = -[25x + 30y]_{(0,0)}^{(2,2)}$.
$V(2, 2) = -[25(2) + 30(2)] - [25(0) + 30(0)]$.
$V(2, 2) = -(50 + 60) = -110 \, V$.

(D) The relation between electric field $\vec{E}$ and electric potential $V$ is given by $\vec{E} = -\frac{dV}{dx} \hat{i}$.
Therefore, $dV = -E_x dx$.
Integrating both sides from $x = -5$ to $x = 1$:
$\int_{V_2}^{V_1} dV = -\int_{-5}^{1} (Ax + B) dx$
$V_1 - V_2 = -\int_{-5}^{1} (20x + 10) dx$
$V_1 - V_2 = -[10x^2 + 10x]_{-5}^{1}$
$V_1 - V_2 = -[(10(1)^2 + 10(1)) - (10(-5)^2 + 10(-5))]$
$V_1 - V_2 = -[(10 + 10) - (250 - 50)]$
$V_1 - V_2 = -[20 - 200]$
$V_1 - V_2 = -[-180] = 180\, V$.

(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 $|\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2}$.
$|\vec{E}| = \sqrt{(5)^2 + (-3)^2 + (-\sqrt{15})^2} = \sqrt{25 + 9 + 15} = \sqrt{49} = 7$.

(A) The relationship between electric potential $V$ and electric field $\overrightarrow{E}$ is given by the formula $\Delta V = -\int \overrightarrow{E} \cdot d \overrightarrow{r}$.
This can be written as $\Delta V = -E \cdot dr \cdot \cos \theta$,where $\theta$ is the angle between the electric field vector $\overrightarrow{E}$ and the displacement vector $d \overrightarrow{r}$.
For the variation in electric potential to be maximum,the magnitude of $\cos \theta$ must be maximum.
The maximum value of $|\cos \theta|$ is $1$,which occurs when $\theta = 0^{\circ}$ or $\theta = 180^{\circ}$.
This corresponds to moving along the direction of the electric field lines (the lines of force).

(D) The electric field $\vec E$ is related to the potential $V$ by the relation $\vec E = -\nabla V$.
Since $V = 2x^2$,the electric field is given by $\vec E = -\left( \frac{\partial V}{\partial x} \hat i + \frac{\partial V}{\partial y} \hat j + \frac{\partial V}{\partial z} \hat k \right)$.
Calculating the partial derivatives: $\frac{\partial V}{\partial x} = 4x$,$\frac{\partial V}{\partial y} = 0$,and $\frac{\partial V}{\partial z} = 0$.
Thus,$\vec E = -(4x) \hat i$.
At the point $(2 \ m, 0, 3 \ m)$,the $x$-coordinate is $2 \ m$.
Substituting $x = 2$,we get $\vec E = -(4 \times 2) \hat i = -8 \hat i \ N C^{-1}$.

(C) 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 $V = \frac{1}{2}(y^2 - 4x)$.
Calculating the partial derivatives:
$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x} \left( \frac{1}{2}y^2 - 2x \right) = -(-2) = 2 \text{ V/m}$.
$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y} \left( \frac{1}{2}y^2 - 2x \right) = -\left( \frac{1}{2} \cdot 2y \right) = -y \text{ V/m}$.
At the point $(1 \text{ m}, 1 \text{ m})$:
$E_x = 2 \text{ V/m}$.
$E_y = -1 \text{ V/m}$.
Therefore,the electric field is $\vec{E} = 2\hat{i} - \hat{j} \text{ V/m}$.

(B) The relation between electric field $\vec{E}$ and potential $V$ is given by $\vec{E} = -\nabla V$.
Given $\vec{E} = a(y\hat{i} + x\hat{j})$,we have:
$-\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}\right) = ay\hat{i} + ax\hat{j}$.
Comparing the components:
$\frac{\partial V}{\partial x} = -ay$ and $\frac{\partial V}{\partial y} = -ax$.
Integrating $\frac{\partial V}{\partial x} = -ay$ with respect to $x$:
$V = -axy + f(y)$,where $f(y)$ is a function of $y$.
Differentiating this with respect to $y$:
$\frac{\partial V}{\partial y} = -ax + f'(y)$.
Comparing this with $\frac{\partial V}{\partial y} = -ax$,we get $f'(y) = 0$,which implies $f(y) = C$ (a constant).
Therefore,$V = -axy + C$.

(B) 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 $V = 2x + 3y - z$,we calculate the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(2x + 3y - z) = 2$
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(2x + 3y - z) = 3$
$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(2x + 3y - z) = -1$
Substituting these values into the expression for $\vec{E}$:
$\vec{E} = -(2\hat{i} + 3\hat{j} - 1\hat{k})$
$\vec{E} = -2\hat{i} - 3\hat{j} + \hat{k}$

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

(B) The relationship between electric field $\vec E$ and electric potential $V$ is given by $\vec E = -\nabla V$,which implies $dV = -\vec E \cdot d\vec r$.
Integrating from the origin $(0, 0, 0)$ to the point $(1, 1, 1)$:
$V(1, 1, 1) - V(0, 0, 0) = -\int_{(0,0,0)}^{(1,1,1)} (2x\hat i + y^2\hat j) \cdot (dx\hat i + dy\hat j + dz\hat k)$.
$V(1, 1, 1) - 2 = -\left[ \int_{0}^{1} 2x \, dx + \int_{0}^{1} y^2 \, dy \right]$.
$V(1, 1, 1) - 2 = -\left[ x^2 \right]_{0}^{1} - \left[ \frac{y^3}{3} \right]_{0}^{1}$.
$V(1, 1, 1) - 2 = -(1 - 0) - (1/3 - 0) = -1 - 1/3 = -4/3$.
$V(1, 1, 1) = 2 - 4/3 = 2/3 \, V$.

(B) The potential difference between two points in a uniform electric field $\vec E$ is given by $\Delta V = -\int \vec E \cdot d\vec l$.
For a uniform electric field,this simplifies to $\Delta V = E \Delta r$,where $\Delta r$ is the displacement along the direction of the electric field.
In the given figure,the path length is $d$ at an angle of $60^o$ with the electric field lines.
The component of the displacement $d$ along the direction of the electric field is $\Delta r = d \cos 60^o$.
Therefore,the potential difference between points $1$ and $2$ is $V = E d \cos 60^o$.

(A) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by $\Delta V = -\int \vec{E} \cdot d\vec{r}$.
Given $\vec{E} = 2x\hat{i} + 3y\hat{j}$ and $V(0,0) = 5\, V$.
Integrating from origin $(0,0)$ to point $(X, Y)$:
$V(X, Y) - V(0,0) = -\int_{(0,0)}^{(X, Y)} (2x\hat{i} + 3y\hat{j}) \cdot (dx\hat{i} + dy\hat{j})$
$V(X, Y) - 5 = -\int_{0}^{X} 2x\, dx - \int_{0}^{Y} 3y\, dy$
$V(X, Y) - 5 = -[x^2]_{0}^{X} - [\frac{3}{2}y^2]_{0}^{Y}$
$V(X, Y) - 5 = -X^2 - \frac{3}{2}Y^2$
$V(X, Y) = -X^2 - \frac{3}{2}Y^2 + 5$.

(D) The electric field $E$ and electric potential $V$ are related by the expression $E = -\frac{dV}{dr}$.
If $V = 0$ at a point,it does not necessarily mean $E = 0$. For example,at the center of a uniformly charged spherical shell,$V$ is constant and non-zero,but $E = 0$. Conversely,between two equal and opposite charges,the potential $V$ is zero at the midpoint,but the electric field $E$ is non-zero.
Similarly,if $E = 0$ at a point,it does not imply $V = 0$. For example,inside a charged conducting sphere,$E = 0$ everywhere,but $V$ is constant and equal to the potential at the surface.
Therefore,if $V = 0$,$E$ may or may not be zero. Similarly,if $E = 0$,$V$ may or may not be zero.

(C) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j})$.
Given $V = \frac{1}{2}(y^2 - 4x)$.
Calculating the partial derivative with respect to $x$:
$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x} [\frac{1}{2}(y^2 - 4x)] = -\frac{1}{2}(-4) = 2 \text{ V/m}$.
Calculating the partial derivative with respect to $y$:
$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y} [\frac{1}{2}(y^2 - 4x)] = -\frac{1}{2}(2y) = -y \text{ V/m}$.
At the point $(1 \text{ m}, 1 \text{ m})$,we have $x = 1$ and $y = 1$.
Substituting $y = 1$ into the expression for $E_y$:
$E_y = -1 \text{ V/m}$.
Thus,the electric field vector is $\vec{E} = E_x\hat{i} + E_y\hat{j} = 2\hat{i} - 1\hat{j} \text{ V/m}$.

(N/A) The electric field $E$ of a point charge $Q$ at a distance $r$ is given by $E = \frac{kQ}{r^2}$.
The electrostatic potential $V$ of a point charge $Q$ at a distance $r$ is given by $V = \frac{kQ}{r}$.
Comparing the two expressions,we can see that $V = E \times r$.
Key differences:
$1$. Electric field is a vector quantity,whereas electrostatic potential is a scalar quantity.
$2$. The electric field represents the force per unit charge,while the electrostatic potential represents the work done per unit charge in bringing a test charge from infinity to a point.

(N/A) The electric field $\vec{E}$ at any point is related to the electrostatic potential $V$ at that point by the negative gradient of the potential.
Mathematically,this is expressed as: $\vec{E} = -\nabla V$.
In Cartesian coordinates,this can be written as: $\vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
This relation signifies that the electric field points in the direction of the steepest decrease of the electrostatic potential.

(N/A) As shown in the figure,consider two closely spaced equipotential surfaces $A$ and $B$ with potential values $V$ and $V+\delta V$,where $\delta V$ is the change in $V$ in the direction of the electric field $\vec{E}$.
Let $P$ be a point on the surface $B$. $\delta l$ is the perpendicular distance of the surface $A$ from $P$. Suppose that a unit positive charge is moved along the perpendicular from the surface $B$ to the surface $A$ against the electric field. The work done in this process is $|\vec{E}| \delta l$.
But work done,$W = V_{A} - V_{B}$.
Therefore,$|\vec{E}| \delta l = V - (V + \delta V)$.
$|\vec{E}| \delta l = -\delta V$.
$|\vec{E}| = -\frac{\delta V}{\delta l}$.
Hence,the negative value of the potential gradient is equal to the magnitude of the electric field. $\frac{\delta V}{\delta l}$ is known as the potential gradient. Its unit is $V \cdot m^{-1}$.
From this,there are two important conclusions:
$(1)$ The electric field is in the direction in which the potential decreases most steeply.
$(2)$ The magnitude of the electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

(N/A) The electric field $E$ at a point is defined as the negative gradient of the electrostatic potential $V$ at that point.
Mathematically,this is expressed as:
$E = -\nabla V$
In one dimension,this relation is given by:
$E = -\frac{dV}{dr}$
where $E$ is the electric field,$V$ is the electrostatic potential,and $r$ is the position coordinate.

(N/A) The potential gradient is defined as the rate of change of electric potential with respect to distance in the direction of the electric field.
Mathematically,it is expressed as $E = -\frac{dV}{dr}$,where $E$ is the electric field intensity,$V$ is the electric potential,and $r$ is the distance.
The negative sign indicates that the electric potential decreases in the direction of the electric field.
The $SI$ unit of potential gradient is $V/m$ (volt per meter).

(B) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by the gradient formula: $\vec{E} = -\nabla V$.
Since the electric potential $V$ is constant $(5 \ V)$ throughout the given region,its spatial derivative (gradient) is zero.
Therefore,$\vec{E} = -\frac{dV}{dr} = 0$.
Thus,the magnitude of the electric field in this region is $0 \ N/C$.

(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 $V = 3x^2$,we calculate the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(3x^2) = 6x$
$\frac{\partial V}{\partial y} = 0$
$\frac{\partial V}{\partial z} = 0$
Thus,$\vec{E} = -(6x) \hat{i} = -6x \hat{i}$.
At the point $(1, 0, 3)$,the $x$-coordinate is $1$. Substituting this into the expression for $\vec{E}$:
$\vec{E} = -6(1) \hat{i} = -6 \hat{i} \, Vm^{-1}$.
The magnitude is $6 \, Vm^{-1}$ and the negative sign indicates it is directed along the negative $x$-axis.

(A) The potential gradient is defined as the rate of change of electric potential with respect to distance.
Mathematically,it is expressed as $\frac{dV}{dr}$.
According to the relation between electric field $E$ and potential $V$,we have $E = -\frac{dV}{dr}$.
Since the electric field $E$ is a vector quantity,the potential gradient $\frac{dV}{dr}$ must also be a vector quantity.
Therefore,the correct option is $A$.

(A) The potential difference $V_A - V_B$ is given by the area under the $E-x$ graph between the points $x_A$ and $x_B$.
$V_2 - V_6 = \int_{2}^{6} E \, dx$
This integral represents the area under the $E-x$ graph from $x = 2 \, m$ to $x = 6 \, m$.
The area consists of a rectangle from $x = 2$ to $x = 4$ and a triangle from $x = 4$ to $x = 6$.
Area of rectangle = $base \times height = (4 - 2) \times 10 = 2 \times 10 = 20 \, V$.
Area of triangle = $\frac{1}{2} \times base \times height = \frac{1}{2} \times (6 - 4) \times 10 = \frac{1}{2} \times 2 \times 10 = 10 \, V$.
Total potential difference = $20 + 10 = 30 \, V$.

(A) The electric field $E$ is perpendicular to the equipotential surfaces and points from higher potential to lower potential.
From the figure,the equipotential lines make an angle of $135^{\circ}$ with the positive $x$-axis.
The normal to these lines makes an angle of $135^{\circ} - 90^{\circ} = 45^{\circ}$ with the positive $x$-axis.
The potential difference between two consecutive surfaces is $\Delta V = 10 \, V$.
The perpendicular distance $d$ between two consecutive surfaces can be calculated from the geometry: $d = 1 \cdot \sin(45^{\circ}) = \frac{1}{\sqrt{2}} \, m$.
The magnitude of the electric field is $E = \frac{|\Delta V|}{d} = \frac{10}{1/\sqrt{2}} = 10 \sqrt{2} \, V/m$.
Since the field points from higher potential to lower potential,it is directed at $45^{\circ}$ with the $x$-axis.

(B) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V$.
Given $V = 10axy$.
The components of the electric field are given by:
$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(10axy) = -10ay$
$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(10axy) = -10ax$
$E_z = -\frac{\partial V}{\partial z} = 0$
Thus,the electric field vector is $\vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k} = -10ay \hat{i} - 10ax \hat{j} = -10a(y \hat{i} + x \hat{j})$.

(B) The electric field component along the $x$-axis is given by $E_x = -\frac{dV}{dx}$.
Given that the potential decreases uniformly from $60 \,V$ to $20 \,V$ over a distance of $4 \,m$ (from $x = -2 \,m$ to $x = +2 \,m$),the magnitude of the $x$-component of the electric field is $|E_x| = \frac{\Delta V}{\Delta x} = \frac{60 \,V - 20 \,V}{2 \,m - (-2 \,m)} = \frac{40 \,V}{4 \,m} = 10 \,V/m$.
Since the potential is only given as a function of $x$,we cannot conclude that the electric field components in the $y$ or $z$ directions are zero. The total electric field is $E = \sqrt{E_x^2 + E_y^2 + E_z^2}$.
Even if $E_y$ and $E_z$ are non-zero,the magnitude of the electric field $E$ will be $\sqrt{(10)^2 + E_y^2 + E_z^2}$,which is greater than or equal to $10 \,V/m$.
Therefore,the magnitude of the electric field at the origin may be greater than $10 \,V/m$.

(B) 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 $V = \frac{3x^2}{2} - \frac{y^2}{4}$.
Calculating the partial derivatives:
$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x} \left( \frac{3x^2}{2} - \frac{y^2}{4} \right) = -3x$.
$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y} \left( \frac{3x^2}{2} - \frac{y^2}{4} \right) = -\left( -\frac{2y}{4} \right) = \frac{y}{2}$.
At the point $(1 \, m, 2 \, m)$,substitute $x = 1$ and $y = 2$:
$E_x = -3(1) = -3 \, N/C$.
$E_y = \frac{2}{2} = 1 \, N/C$.
Thus,the electric field vector is $\vec{E} = -3 \hat{i} + \hat{j} \, N/C$.

(D) The relationship between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.
Given the potential function $V = 8x^2 + 2$.
Differentiating with respect to $x$,we get $\frac{dV}{dx} = \frac{d}{dx}(8x^2 + 2) = 16x$.
Thus,the electric field $E_x = -16x$.
At the point $(-4, 0)$,the $x$-coordinate is $x = -4$.
Substituting this value into the expression for $E_x$,we get $E_x = -16(-4) = 64 \, V/m$.
The magnitude of the electric field is $|E_x| = 64 \, V/m$.

(B) The potential difference between points $B$ and $C$ is $\Delta V = V_C - V_B = 20 \, V - 15 \, V = 5 \, V$.
Since the electric field $\vec{E}$ is uniform and in the plane of the triangle,the potential difference is given by $\Delta V = E \cdot d \cos \theta$,where $d$ is the distance between the points and $\theta$ is the angle between the electric field and the displacement vector.
Since $A$ and $B$ are at the same potential,the line $AB$ is an equipotential line. The electric field $\vec{E}$ must be perpendicular to the equipotential line $AB$. Thus,$\vec{E}$ is parallel to $BC$.
The distance between $B$ and $C$ is $d = BC = 4 \, cm = 0.04 \, m$.
Since $\vec{E}$ is parallel to $BC$,the potential difference is $\Delta V = E \cdot d$.
$5 \, V = E \times 0.04 \, m$.
$E = \frac{5}{0.04} \, V/m = 125 \, V/m$ (or $N/C$).

(B) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.
Given $V = 2ar^2 + b$,differentiating with respect to $r$ gives $E = -\frac{d}{dr}(2ar^2 + b) = -4ar$.
According to Gauss's Law for a uniformly charged sphere,the electric field inside is given by $E = \frac{\rho r}{3\varepsilon}$,where $\rho$ is the volume charge density.
Comparing the two expressions for $E$: $-4ar = \frac{\rho r}{3\varepsilon}$.
Solving for $\rho$: $\rho = -12a\varepsilon$.
Comparing this with the given form $\rho = -\lambda a\varepsilon$,we find $\lambda = 12$.

(B) The electric field $E$ between two large parallel conducting plates is uniform and is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the separation between the plates.
Given $d = 10 \ cm$,so $E = \frac{V}{10 \ cm}$.
The potential difference between two points is given by $\Delta V = E \cdot \Delta x$,where $\Delta x$ is the projection of the displacement vector between the points along the direction of the electric field.
The electric field is directed from the positive plate to the negative plate (horizontally).
The horizontal distance (projection) between points $A$ and $B$ is the same as the horizontal distance between $C$ and $B$,which is $4 \ cm$.
Therefore,the potential difference between $A$ and $B$ is $V_{AB} = E \cdot (4 \ cm) = \left( \frac{V}{10 \ cm} \right) \times 4 \ cm = \frac{4}{10} V = \frac{2}{5} V$.

(D) The electric field $\vec{E}$ 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:
$\frac{\partial V}{\partial x} = 6 - 8y^2$
$\frac{\partial V}{\partial y} = -16xy - 8 + 6z$
$\frac{\partial V}{\partial z} = 6y - 8z$
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,$\vec{E} = -(6 \hat{i} - 8 \hat{j} + 0 \hat{k}) = -6 \hat{i} + 8 \hat{j} \text{ V/m}$.
The magnitude of the electric field is $|\vec{E}| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N/C}$.
The electric force is $F = qE = 2 \text{ C} \times 10 \text{ N/C} = 20 \text{ N}$.

(A) The electric field $E$ is related to the potential difference $\Delta V$ and the perpendicular distance $d$ between equipotential surfaces by the formula $E = \frac{\Delta V}{d}$.
From the figure,the potential difference between two adjacent equipotential surfaces is $\Delta V = 20 \ V - 10 \ V = 10 \ V$.
The distance along the $x$-axis between these two surfaces is $\Delta x = 20 \ cm - 10 \ cm = 10 \ cm = 0.1 \ m$.
The perpendicular distance $d$ between the surfaces is given by $d = \Delta x \sin 30^{\circ} = 0.1 \times \sin 30^{\circ} = 0.1 \times 0.5 = 0.05 \ m$.
Therefore,the magnitude of the electric field is $E = \frac{10 \ V}{0.05 \ m} = 200 \ V/m$.

(D) The relation between electric field $E$ and electric potential $V$ is given by $E = -\frac{dV}{dx}$.
Given $V = 2x^2 + 10x - 9$.
Differentiating $V$ with respect to $x$:
$\frac{dV}{dx} = \frac{d}{dx}(2x^2 + 10x - 9) = 4x + 10$.
Therefore,$E = -(4x + 10)$.
At $x = 1 \text{ m}$,the electric field is:
$E = -(4(1) + 10) = -(4 + 10) = -14 \text{ V/m}$.

(A) For the particle to be in equilibrium,the upward electric force must balance the downward gravitational force.
$F_e = mg$
Since $F_e = qE$ and $E = \frac{V}{d}$,we have $F_e = \frac{qV}{d}$.
Equating the forces: $mg = \frac{qV}{d}$.
Rearranging for mass $m$: $m = \frac{qV}{gd}$.
Given values: $q = 1.6 \times 10^{-19} \text{ C}$,$V = 980 \text{ V}$,$d = 8 \text{ cm} = 0.08 \text{ m}$,$g = 9.8 \text{ m/s}^2$.
Substituting the values:
$m = \frac{1.6 \times 10^{-19} \times 980}{9.8 \times 0.08}$
$m = \frac{1.6 \times 10^{-19} \times 100}{0.08}$
$m = \frac{1.6 \times 10^{-17}}{0.08} = 20 \times 10^{-17} \text{ kg} = 2 \times 10^{-16} \text{ kg}$.

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

(B) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.
Given $V = 4x^2 + 8x - 3$.
Differentiating $V$ with respect to $x$:
$\frac{dV}{dx} = \frac{d}{dx}(4x^2 + 8x - 3) = 8x + 8$.
Substituting this into the electric field formula:
$E = -(8x + 8) = -8x - 8$.
Now,evaluate the electric field at $x = 0.5 \ m$:
$E = -8(0.5) - 8 = -4 - 8 = -12 \ V/m$.
Therefore,the correct option is $B$.

(A) The electric force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$.
Thus, the electric field strength is $E = F/q = 3000 \, N / 3 \, C = 1000 \, N/C$.
The potential difference $V$ between two points separated by a distance $d$ in a uniform electric field is given by $V = Ed$.
Given $d = 1 \, cm = 10^{-2} \, m$.
Substituting the values, $V = 1000 \, N/C \times 10^{-2} \, m = 10 \, V$.

(A) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by the relation: $dV = -\vec{E} \cdot d\vec{r}$.
Given $\vec{E} = 30x^2 \hat{i}$ and $d\vec{r} = dx \hat{i}$,we have $dV = -30x^2 dx$.
To find the potential difference $(V_A - V_0)$,we integrate from $x = 0$ to $x = 2$:
$\int_{V_0}^{V_A} dV = -\int_{0}^{2} 30x^2 dx$.
$V_A - V_0 = -[10x^3]_{0}^{2}$.
$V_A - V_0 = -(10(2)^3 - 10(0)^3) = -(10 \times 8) = -80 \ V$.

(C) The relationship between electric field $\vec{E}$ and electric potential $V$ is given by the equation: $dV = -\vec{E} \cdot d\vec{l}$.
Given $\vec{E} = 20x^2 \hat{i}$ and $d\vec{l} = dx \hat{i}$.
Substituting these into the equation,we get: $dV = -(20x^2 \hat{i}) \cdot (dx \hat{i}) = -20x^2 dx$.
To find the potential difference $V_A - V_0$,we integrate from $x=0$ to $x=3 \ m$:
$V_A - V_0 = \int_{V_0}^{V_A} dV = \int_{0}^{3} -20x^2 dx$.
$V_A - V_0 = -20 \left[ \frac{x^3}{3} \right]_{0}^{3}$.
$V_A - V_0 = -20 \left( \frac{3^3}{3} - 0 \right) = -20 \left( \frac{27}{3} \right) = -20 \times 9 = -180 \ V$.

(D) The relation between electric field $\overrightarrow{E}$ and electric potential $V$ is given by $\overrightarrow{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$.
Given $V = 5x^2$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{d}{dx}(5x^2) = 10x$
$\frac{\partial V}{\partial y} = 0$
$\frac{\partial V}{\partial z} = 0$
Thus,$\overrightarrow{E} = -(10x)\hat{i} = -10x\hat{i} \text{ N/C}$.
At the point $(1, 2, 3) \text{ m}$,substituting $x = 1$:
$\overrightarrow{E} = -10(1)\hat{i} = -10\hat{i} \text{ N/C}$.
Therefore,the correct option is $D$.

(C) The electric field $E$ between two parallel plates is given by the formula $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates.
Given:
Potential difference $V = 10 \ V$
Distance $d = 20 \ cm = 20 \times 10^{-2} \ m = 0.2 \ m$
Substituting these values into the formula:
$E = \frac{10}{0.2} = 50 \ Vm^{-1}$
Therefore,the electric field between the plates is $50 \ Vm^{-1}$.

(B) 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 distance between the plates.
Given:
Potential difference $\Delta V = V_2 - V_1 = 30 \,V - (-10 \,V) = 40 \,V$.
Distance $d = 2 \,cm = 2 \times 10^{-2} \,m$.
Substituting the values:
$E = \frac{40 \,V}{2 \times 10^{-2} \,m} = 20 \times 10^2 \,V/m = 2000 \,V/m$.
Therefore,the correct option is $B$.