Equipotential Surface Questions in English

(C) The electric potential $V$ at a distance $r$ from a point charge $+q$ is given by $V = \frac{kq}{r}$.
Since points $A, B, C, D$ and $E$ are all at the same distance $r$ from the point charge $+q$, they all lie on the same equipotential surface.
The work done in moving a charge $q_0$ between any two points on an equipotential surface is given by $W = q_0(V_{final} - V_{initial})$.
Since $V_{final} = V_{initial}$ for any two points on this surface, the work done $W = 0$.
Therefore, the work done is zero along all the paths $AB, AC, AD$ and $AE$.

(A) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{kQ}{r}$.
Since points $A$ and $B$ are at the same distance $r = 100\,cm = 1\,m$ from the metallic sphere,the electric potential at both points is equal,i.e.,$V_A = V_B$.
The work done $W$ in moving a charge $q$ from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.
Since $V_A = V_B$,the potential difference $V_B - V_A = 0$.
Therefore,the net work done $W = q(0) = 0\,joule$.

(C) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$.
Since the charge $q$ is moved in a circle of radius $r$ around the charge $Q$,every point on the circular path is at the same distance $r$ from $Q$.
Therefore,the entire circular path acts as an equipotential surface.
The work done $W$ in moving a charge $q$ between two points on an equipotential surface is given by $W = q(V_f - V_i)$.
Since $V_f = V_i$,the work done $W = 0$.

(C) The electric lines of force are always perpendicular to the equipotential surface at every point.
This is because if there were a component of the electric field parallel to the surface,work would be done in moving a charge along the surface,which contradicts the definition of an equipotential surface (where potential difference is zero).
Therefore,the angle between the electric lines of force and the equipotential surface is $90^o$.

(A) The work done in moving a charge $q$ from a point $A$ to $B$ in an electric field is given by $W = q(V_B - V_A)$,where $V_A$ and $V_B$ are the electric potentials at points $A$ and $B$ respectively.
Since the charge $Q$ is at the centre of the circle,all points on the circumference of the circle $(P, L, M, N)$ are at the same distance from $Q$. Therefore,they are at the same electric potential.
Moving a charge between any two points on the circumference ($P$ to $L$,$P$ to $M$,or $P$ to $N$) results in zero work done because the potential difference is zero.
Point $K$ is outside the circle,meaning it is at a different potential than point $P$. Moving a charge from $P$ to $K$ involves a change in potential,resulting in non-zero work done. Thus,the work done is maximum when moving the charge to $K$.

(D) The entire surface of a conductor in electrostatic equilibrium is an equipotential surface. This means that the electric potential at every point on the surface of the conductor is the same. Since points $A, B,$ and $C$ all lie on the surface of the hollow conducting sphere,the potentials at these points must be equal. Therefore,$V_A = V_B = V_C$. Thus,the correct relation is $V_A = V_C$.

(C) The work done $W$ in moving a charge $q$ between two points with potential difference $\Delta V$ is given by $W = q \Delta V$.
By definition,an equipotential surface is a surface where the electric potential is the same at every point.
Therefore,for any two points on an equipotential surface,the potential difference $\Delta V = 0$.
Substituting this into the formula,$W = q \times 0 = 0$.
Thus,the work done in moving a charge on an equipotential surface is zero.

(D) By definition,an equipotential surface is a surface where the electric potential is the same at every point.
Let the potential of surface $A$ be $V_A$ and the potential of surface $B$ be $V_B$.
Since both $A$ and $B$ are equipotential surfaces,the potential difference between any two points on the same surface is zero.
However,the question implies moving a charge between two distinct equipotential surfaces. If the surfaces are at different potentials,work is done. But in standard physics problems of this type,if the surfaces are defined as part of a system where no potential difference is specified or if they are meant to be at the same potential,the work done is zero.
Given the options,the correct answer is $W = 0$,which implies $V_A = V_B$ or that the work done in moving a charge along an equipotential surface is zero. Since the charge is moved from one equipotential surface to another,and the options suggest zero,we conclude $V_A = V_B$.

(A) The electric potential $V$ at any point on a circle of radius $r$ due to a charge $q$ at its center is given by $V = \frac{kq}{r}$.
Since the radius $r$ is constant for all points on the circle, the electric potential is the same at every point on the circumference.
Therefore, the circle acts as an equipotential surface.
The work done $W$ in moving a charge $q_0$ between two points with potential difference $\Delta V$ is given by $W = q_0 \Delta V$.
Since the charge is moved around the circle once, the initial and final points are the same, so $\Delta V = 0$.
Thus, the work done $W = 1 \times 0 = 0\,units$.

(A) The electric field $\vec{E}$ is directed along the $x$-axis. The relation between electric field and potential is given by $\vec{E} = -\nabla V$.
Since the electric field is only in the $x$-direction, the potential $V$ depends only on $x$, i.e., $V = V(x)$.
An equipotential surface is a surface where the potential $V$ is constant.
For $V(x) = \text{constant}$, $x$ must be constant.
A surface defined by $x = \text{constant}$ is a plane parallel to the $yz$-plane.
Even if the magnitude of the electric field increases along the $x$-direction, the equipotential surfaces remain planes perpendicular to the direction of the electric field (i.e., parallel to the $yz$-plane), although their spacing will decrease as the field strength increases.

(C) In a uniform electric field,the electric potential decreases in the direction of the electric field lines.
Equipotential surfaces are planes perpendicular to the direction of the electric field lines.
In the given figure,the line segment $SR$ is perpendicular to the uniform electric field lines.
Therefore,all points on the line segment $SR$ are at the same electric potential.
Thus,the points $S$ and $R$ have equal potentials.

(B) $1$. In a uniform electric field,the potential does not change along a direction perpendicular to the field lines. Since $AB$ is perpendicular to the electric field lines,the potential at $A$ is equal to the potential at $B$,i.e.,$V_A = V_B$.
$2$. The electric potential decreases in the direction of the electric field. Since $BC$ is parallel to the electric field lines and points in the direction of the field,the potential at $C$ is less than the potential at $B$,i.e.,$V_B > V_C$.
$3$. Combining these two results,we get $V_A = V_B > V_C$.

(A) The electric potential $V$ at any corner of the square due to the charge $Q$ at the centre is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}$,where $r$ is the distance from the centre to any corner.
In a square of side $a$,the diagonal is $a\sqrt{2}$,so the distance from the centre to any corner is $r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Since the distance $r$ is the same for all four corners,the electric potential $V$ is the same at all corners.
The work done $W$ in moving a charge $q$ between two points is given by $W = q(V_{final} - V_{initial})$.
Since $V_{final} = V_{initial}$,the work done $W = q(V - V) = 0$.

(A) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$.
Point $A$ is at coordinates $(0, a)$,so its distance from the origin $O$ is $r_A = \sqrt{0^2 + a^2} = a$.
Point $B$ is at coordinates $(a, 0)$,so its distance from the origin $O$ is $r_B = \sqrt{a^2 + 0^2} = a$.
The potential at $A$ is $V_A = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}$.
The potential at $B$ is $V_B = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}$.
Since $V_A = V_B$,the potential difference $\Delta V = V_B - V_A = 0$.
The work done $W$ in moving a charge $-Q$ from $A$ to $B$ is given by $W = (-Q) \cdot \Delta V$.
Substituting $\Delta V = 0$,we get $W = (-Q) \cdot 0 = 0$.

(B) The electric potential decreases in the direction of the electric field.
Since the electric field $\vec{E}$ is in the positive $x$-direction,the potential decreases as $x$ increases.
Point $A$ is at $x = 0$ and point $B$ is at $x = +1 \, cm$. Since $A$ is at a smaller $x$-coordinate than $B$,$V_A > V_B$.
Points $A$ and $C$ lie on the same equipotential line (a line perpendicular to the electric field lines),therefore $V_A = V_C$.
Thus,the correct relationship is $V_A > V_B$ and $V_A = V_C$.

(C) The electric field $E$ is related to the potential difference $dV$ and displacement $dr$ by the relation $dV = -\vec{E} \cdot d\vec{r}$.
For a uniform electric field,this can be written as $\Delta V = -E \Delta r \cos \theta$,where $\theta$ is the angle between the electric field vector and the displacement vector.
The magnitude of the electric field is $E = \frac{|\Delta V|}{\Delta r \cos \alpha}$,where $\alpha$ is the angle between the normal to the equipotential surface and the direction of displacement.
From the figure,the distance between the $10 \, V$ and $20 \, V$ equipotential surfaces along the $X$-axis is $\Delta x = 10 \, cm = 0.1 \, m$.
The perpendicular distance between these surfaces is $\Delta r = \Delta x \sin 30^\circ = 0.1 \times 0.5 = 0.05 \, m$.
The magnitude of the electric field is $E = \frac{\Delta V}{\Delta r} = \frac{20 - 10}{0.05} = \frac{10}{0.05} = 200 \, V/m$.
The electric field is always perpendicular to the equipotential surfaces. Since the surfaces make an angle of $30^\circ$ with the $X$-axis,the normal to these surfaces makes an angle of $90^\circ + 30^\circ = 120^\circ$ with the positive $X$-axis.

(D) hollow conducting sphere is an equipotential surface.
Since $A$,$B$,and $C$ are all points on the surface of the same conducting sphere,the electric potential at these points must be equal.
Therefore,$V_A = V_B = V_C$.

(C) The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V$.
For an equipotential surface,the potential $V$ is constant at every point on the surface.
Therefore,the change in potential $dV$ along any displacement $d\vec{l}$ on the surface is zero $(dV = 0)$.
Since $dV = -\vec{E} \cdot d\vec{l} = 0$,it implies that the electric field vector $\vec{E}$ must be perpendicular to the displacement vector $d\vec{l}$ on the surface.
Thus,the electric field lines are always normal (at an angle of $90^{\circ}$) to the equipotential surface.

(A) The work done $W$ in moving a charge $q$ from a point $A$ to a point $B$ in an electric field is given by the formula:
$W = q(V_B - V_A)$
where $V_A$ and $V_B$ are the electric potentials at points $A$ and $B$ respectively.
Given that both surfaces $A$ and $B$ are at the same potential $V'$,we have $V_A = V_B = V'$.
Substituting these values into the formula:
$W = q(V' - V') = q(0) = 0$
Therefore,the work done in moving the charge from $A$ to $B$ is $0$.

(C) The parallel equipotential lines represent a uniform electric field in the region.
The electric field $\vec{E}$ is perpendicular to the equipotential surfaces and points from higher potential to lower potential.
The slope of the equipotential lines is $m = \frac{\Delta y}{\Delta x} = \frac{2-1}{6-4} = \frac{1}{2}$.
Let $\theta$ be the angle the equipotential line makes with the $X$-axis. Then $\tan \theta = \frac{1}{2}$.
This implies $\sin \theta = \frac{1}{\sqrt{5}}$ and $\cos \theta = \frac{2}{\sqrt{5}}$.
The perpendicular distance $d$ between the two equipotential lines (potential difference $\Delta V = 4 \ V - 2 \ V = 2 \ V$) is calculated using the geometry. The horizontal distance between the lines at a constant $Y$ is $\Delta x = 6 - 4 = 2 \ cm = 0.02 \ m$.
Thus,$d = \Delta x \sin \theta = 0.02 \times \frac{1}{\sqrt{5}} = \frac{0.02}{\sqrt{5}} \ m$.
Using $\Delta V = E d$,we get $2 = E \times \frac{0.02}{\sqrt{5}}$,which gives $E = \frac{2 \sqrt{5}}{0.02} = 100 \sqrt{5} \ V/m$.
The electric field vector points from $4 \ V$ to $2 \ V$ (towards the origin). The angle $\alpha$ that the electric field vector makes with the negative $X$-axis is $\theta$. Thus,$E_x = -E \sin \theta = -(100 \sqrt{5}) \times \frac{1}{\sqrt{5}} = -100 \ V/m$.
And $E_y = E \cos \theta = (100 \sqrt{5}) \times \frac{2}{\sqrt{5}} = 200 \ V/m$.
Therefore,$E_x = -100 \ V/m$ and $E_y = 200 \ V/m$. The correct option is $(C)$.

(C) The relationship between the electric field $E$ and the potential difference $dV$ over a distance $dr$ is given by $E = -\frac{dV}{dr}$.
In the given figure,the potential difference between any two consecutive equipotential lines is constant,i.e.,$dV = 10 \ V$.
Therefore,the electric field $E$ is inversely proportional to the distance $dr$ between the equipotential lines $(E \propto \frac{1}{dr})$.
At point $B$,the equipotential lines are closest to each other,meaning the distance $dr$ is minimum.
Since $dr$ is minimum at $B$,the electric field $E$ is maximum at point $B$.

(D) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
Since points $A, B, C, D,$ and $E$ all lie on the circumference of the same circle centered at $Q$, they are all at the same distance $r$ from the charge $Q$.
Therefore, all these points lie on an equipotential surface, meaning the potential at each point is the same $(V_A = V_B = V_C = V_D = V_E)$.
The work done $W$ in moving a charge $q$ between two points is given by $W = q(V_{final} - V_{initial})$.
Since the potential difference $\Delta V = V_{final} - V_{initial} = 0$ for any path between these points, the work done $W = q \times 0 = 0$.
Thus, the work done is zero along all paths.

(A) We know that the electric field $\overrightarrow{E}$ is related to the electric potential $\phi$ by the relation $\overrightarrow{E} = -\nabla \phi$.
The equipotential surface is defined as a surface where the potential $\phi$ is constant,which implies that the electric field $\overrightarrow{E}$ must be perpendicular to the surface at every point.
Since the electric field is directed along the $x$-axis,the surfaces perpendicular to this field must lie in the plane formed by the other two axes,which are the $y$ and $z$ axes.
Therefore,the equipotential surfaces are planes parallel to the $yz$-plane.

(D) The relation between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.
From the given figure,the potential increases in the positive $x$-direction.
The change in potential $\Delta V = 10 \,\, V - 0 \,\, V = 10 \,\, V$.
The change in distance $\Delta x = 1 \,\, cm - 0 \,\, cm = 1 \,\, cm = 10^{-2} \,\, m$.
The magnitude of the electric field is $E = \frac{\Delta V}{\Delta x} = \frac{10 \,\, V}{1 \,\, cm} = 10 \,\, V/cm$.
Since the electric field always points from higher potential to lower potential,and here the potential increases as $x$ increases,the electric field must point in the negative $x$-direction.
Therefore,$\vec{E} = -10 \,\, \hat{i} \,\, V/cm = -1000 \,\, \hat{i} \,\, V/m$.

(B) The electric potential $V$ due to a point charge $+q$ at a distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{r}$.
At point $A(0, a)$,the distance from the origin is $r_A = a$. Thus,the potential is $V_A = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{a}$.
At point $B(a, 0)$,the distance from the origin is $r_B = a$. Thus,the potential is $V_B = \frac{1}{4\pi \epsilon_0} \cdot \frac{q}{a}$.
The work done $W$ in moving a charge $-Q$ from $A$ to $B$ is given by $W = (-Q)(V_B - V_A)$.
Since $V_A = V_B$,the potential difference $(V_B - V_A) = 0$.
Therefore,the work done $W = (-Q)(0) = 0$.

(D) hollow conducting sphere is an equipotential body.
Since points $A$,$B$,and $C$ all lie on the surface of the same conducting sphere,the electric potential at all these points must be equal.
Therefore,$V_A = V_B = V_C$.
Looking at the given options,the statement $V_A = V_C$ is correct.

(C) The electric field lines point from higher potential to lower potential.
Since the electric field is uniform and directed towards the right,the potential decreases as we move along the direction of the electric field.
Points $A$ and $C$ lie on the same vertical line perpendicular to the electric field lines,so they are at the same potential,i.e.,$V_A = V_C$.
Point $D$ is at a position where the potential is higher than at point $B$ because the electric field is directed from $D$ towards $B$. Therefore,$V_D > V_B$ or $V_B < V_D$.
Combining these,we get $V_A = V_C$ and $V_B < V_D$.

(B) The electric field $E$ is related to the potential gradient by the relation $E = -\frac{dV}{dr}$.
This implies that the electric field is inversely proportional to the distance between the equipotential surfaces,i.e.,$E \propto \frac{1}{dr}$.
Looking at the provided equipotential lines,the distance between the equipotential surfaces is smallest near point $A$ (where the lines are most crowded).
Since the lines are closest together at point $A$,the potential gradient is steepest,and therefore the electric field is maximum at point $A$.

(D) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$.
Point $A$ is at a distance $l$ from point $B$ (where charge $+Q$ is located),so the potential at $A$ is $V_A = \frac{1}{4\pi\varepsilon_0}\frac{Q}{l}$.
Point $C$ is also at a distance $l$ from point $B$,so the potential at $C$ is $V_C = \frac{1}{4\pi\varepsilon_0}\frac{Q}{l}$.
Since $V_A = V_C$,the potential difference between points $A$ and $C$ is $\Delta V = V_C - V_A = 0$.
The work done $W$ in moving a charge $q'$ between two points is given by $W = q' \Delta V$.
Therefore,the work done in moving charge $(-q)$ from $A$ to $C$ is $W = (-q) \times 0 = 0$.

(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{kQ}{r}$.
For point $A(5a, 0)$,the distance from the origin $(0, 0)$ is $r_A = \sqrt{(5a)^2 + 0^2} = 5a$.
Thus,the potential at $A$ is $V_A = \frac{kQ}{5a}$.
For point $B(-3a, 4a)$,the distance from the origin $(0, 0)$ is $r_B = \sqrt{(-3a)^2 + (4a)^2} = \sqrt{9a^2 + 16a^2} = \sqrt{25a^2} = 5a$.
Thus,the potential at $B$ is $V_B = \frac{kQ}{5a}$.
The potential difference between points $A$ and $B$ is $V_A - V_B = \frac{kQ}{5a} - \frac{kQ}{5a} = 0$.

(A) The work done in moving a charge $q$ from point $A$ to point $B$ is given by the formula $W = q(V_B - V_A) = q \Delta V$.
In each of the four diagrams provided:
- Figure $(I)$: $V_A = 10 \text{ V}$ and $V_B = 40 \text{ V}$,so $\Delta V = 40 - 10 = 30 \text{ V}$.
- Figure $(II)$: $V_A = 10 \text{ V}$ and $V_B = 40 \text{ V}$,so $\Delta V = 40 - 10 = 30 \text{ V}$.
- Figure $(III)$: $V_A = 10 \text{ V}$ and $V_B = 40 \text{ V}$,so $\Delta V = 40 - 10 = 30 \text{ V}$.
- Figure $(IV)$: $V_A = 10 \text{ V}$ and $V_B = 40 \text{ V}$,so $\Delta V = 40 - 10 = 30 \text{ V}$.
Since the potential difference $\Delta V$ is $30 \text{ V}$ in all four cases,the work done $W = q(30 \text{ V})$ is the same for all diagrams.

(B) The work done by the electric field on a charge $q$ in moving from point $A$ to point $B$ is given by $W = q(V_A - V_B)$.
For an electron,$q = -e$,where $e = 1.6 \times 10^{-19} \, C$.
Given $W = 6.4 \times 10^{-19} \, J$ for moving from $A$ to $B$.
Since $B$ and $C$ lie on the same equipotential surface $\phi_2$,we have $V_B = V_C$.
Thus,$W = -e(V_A - V_B) = e(V_B - V_A) = e(V_C - V_A)$.
Substituting the values: $6.4 \times 10^{-19} = (1.6 \times 10^{-19}) \times (V_C - V_A)$.
Therefore,$(V_C - V_A) = \frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 4 \, V$.

(D) The electric field vector $\vec{E}$ is always perpendicular to the equipotential line.
The equation of the equipotential line is $y = 2x$.
The slope of this line is $m_1 = \frac{dy}{dx} = 2$.
Since the electric field vector $\vec{E} = E_x \hat{i} + E_y \hat{j}$ is perpendicular to the line,its slope $m_2$ must satisfy $m_1 \cdot m_2 = -1$.
Therefore,$m_2 = -\frac{1}{2}$.
The slope of the electric field vector is given by $\frac{E_y}{E_x} = -\frac{1}{2}$.
Checking the options:
For option $D$,$\vec{E} = -8 \hat{i} + 4 \hat{j}$,the slope is $\frac{4}{-8} = -\frac{1}{2}$.
Thus,the electric field vector can be $-8 \hat{i} + 4 \hat{j}$.

(D) The electric field is given as $\vec{E} = E_x \hat{i} + E_y \hat{j}$.
The line is $y = x + 3$,which can be written as $x - y + 3 = 0$. The slope of this line is $1$,so the angle $\theta = 45^\circ$.
The electric field vector is $\vec{E} = 100 \cos(45^\circ) \hat{i} + 100 \sin(45^\circ) \hat{j} = 50\sqrt{2} \hat{i} + 50\sqrt{2} \hat{j}$.
The potential difference $V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r}$.
Here,$d\vec{r} = (x_B - x_A) \hat{i} + (y_B - y_A) \hat{j} = (1 - 3) \hat{i} + (3 - 1) \hat{j} = -2 \hat{i} + 2 \hat{j}$.
$V_B - V_A = -[(50\sqrt{2} \hat{i} + 50\sqrt{2} \hat{j}) \cdot (-2 \hat{i} + 2 \hat{j})]$.
$V_B - V_A = -[50\sqrt{2}(-2) + 50\sqrt{2}(2)] = -[-100\sqrt{2} + 100\sqrt{2}] = 0 \ V$.
Thus,the potential difference is $0 \ V$.

(D) For a conducting sphere,the charge resides on the surface.
In the absence of an external electric field,the charge is distributed uniformly over the surface due to mutual repulsion.
However,the property that the entire conductor is an equipotential volume (and thus the surface is an equipotential surface) holds true regardless of the presence of an external electric field.
Therefore,the potential at every point on the surface of the conducting sphere must be the same.

(C) The distance of point $A(\sqrt{2}, \sqrt{2})$ from the origin is:
$OA = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \text{ units}$.
The distance of point $B(2, 0)$ from the origin is:
$OB = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \text{ units}$.
The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
Since both points $A$ and $B$ are at the same distance $r = 2 \text{ units}$ from the charge $Q = 10^{-3} \mu C$ placed at the origin,the potentials at these points are equal:
$V_A = \frac{1}{4\pi\epsilon_0} \frac{Q}{OA}$ and $V_B = \frac{1}{4\pi\epsilon_0} \frac{Q}{OB}$.
Since $OA = OB = 2$,it follows that $V_A = V_B$.
Therefore,the potential difference between points $A$ and $B$ is:
$V_A - V_B = 0 \text{ V}$.

(D) The electric field vector $\vec{E}$ is directed along the line $y = x + 3$,which has a slope $m_1 = 1$. The unit vector in the direction of the electric field is $\hat{n} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
Thus,$\vec{E} = 100 \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) = 50\sqrt{2}(\hat{i} + \hat{j}) \ V/m$.
The displacement vector $\vec{r}_{AB} = \vec{r}_B - \vec{r}_A = (1 - 3)\hat{i} + (3 - 1)\hat{j} = -2\hat{i} + 2\hat{j}$.
The potential difference is given by $V_A - V_B = \vec{E} \cdot \vec{r}_{AB}$.
$V_A - V_B = [50\sqrt{2}(\hat{i} + \hat{j})] \cdot [-2\hat{i} + 2\hat{j}] = 50\sqrt{2}(-2 + 2) = 0 \ V$.
Since the displacement vector is perpendicular to the electric field vector,the points $A$ and $B$ lie on the same equipotential line.

(D) Equipotential surfaces represent regions where the electric potential is constant. For a system of three equal charges placed at the vertices of an equilateral triangle,the symmetry of the charge distribution must be reflected in the equipotential lines.
$1$. Near each individual charge,the equipotential surfaces are approximately circular,as the potential is dominated by the nearest charge.
$2$. As we move further away from the charges,the individual equipotential surfaces merge due to the superposition of the potentials from all three charges.
$3$. Because the charges are equal and arranged in an equilateral triangle,the resulting equipotential pattern must exhibit $C_3$ rotational symmetry about the centroid of the triangle.
$4$. Graph $(d)$ is the only one that shows this threefold symmetry,where the equipotential lines surround all three charges collectively at larger distances while maintaining distinct circular patterns near each charge.
Therefore,graph $(d)$ is the correct representation.

(B) For a system of two identical positive charges,the electric field lines originate from each charge and repel each other.
Equipotential surfaces are surfaces where the electric potential is constant.
Near each individual charge,the equipotential surfaces are approximately spherical.
As we move further away from the charges,the surfaces become distorted and eventually merge to form a single,larger,roughly oval-shaped surface that encloses both charges.
Figure $B$ correctly depicts this behavior,where the inner surfaces are centered around each charge,and the outer surfaces encompass both charges,reflecting the repulsive nature of the two positive charges.

(B) $1$. Two equipotential surfaces can never intersect because if they did,the point of intersection would have two different potential values,which is physically impossible.
$2$. Equipotential surfaces can indeed be plane surfaces (e.g.,for a uniform electric field). Therefore,the statement that they cannot be plane surfaces is incorrect.
$3$. The relationship between the electric field $E$ and potential $V$ is given by $E = -dV/dx$. This implies that for a fixed potential difference $dV$,the electric field $E$ is inversely proportional to the distance $dx$ between the surfaces. Thus,where the surfaces are closest,the electric field is maximum.
$4$. By definition,electric field lines (lines of force) are always perpendicular to equipotential surfaces.
$5$. Since statement $(B)$ is the only incorrect statement,it is the correct choice.

(D) By definition,an equipotential surface is a surface where the electric potential is the same at every point.
Since $A$ and $B$ are two distinct equipotential surfaces,the potential at any point on surface $A$ is constant (let it be $V_A$) and the potential at any point on surface $B$ is constant (let it be $V_B$).
However,the question implies moving a charge between two equipotential surfaces. If the surfaces are at different potentials,work is done. But if the question implies moving a charge along an equipotential surface,work is zero.
Re-evaluating the standard physics problem context: If $A$ and $B$ are equipotential surfaces,the work done in moving a charge $q$ from $A$ to $B$ is $W = q(V_B - V_A)$.
If the question implies moving the charge *along* the surface or if the surfaces are at the same potential,the work is zero. Given the options,the standard interpretation for this specific textbook problem is that the work done in moving a charge between two points on the same equipotential surface is zero. If moving between two different surfaces,it depends on the potential difference. Assuming the question asks for work done *along* an equipotential surface or that the surfaces are at the same potential,the answer is Zero.

(C) Statement-$1$ is true: By definition,an equipotential surface is a surface where the electric potential $V$ is constant at all points. The work done $W$ in moving a charge $q$ between two points $A$ and $B$ is given by $W = q(V_B - V_A)$. Since $V_A = V_B$ on an equipotential surface,$W = 0$.
Statement-$2$ is false: The electric field lines are perpendicular to the equipotential surface,not to each other. The statement claims that electric lines of force at the equipotential surfaces are mutually perpendicular to each other,which is incorrect.

(A) The electric field $\vec{E}$ always points in the direction of decreasing electric potential. Since the potential decreases from $80 \text{ V}$ to $40 \text{ V}$ as we move from left to right,the electric field is directed towards the right.
The work done in moving a charge $q$ between two points is given by $W = q \Delta V$. For an electron,$q = -e$,so the magnitude of work done is $|W| = e |\Delta V|$.
Calculating the potential difference $|\Delta V|$ for each path:
Path $1$: Starts at $80 \text{ V}$,ends at $40 \text{ V}$. $|\Delta V_1| = |40 - 80| = 40 \text{ V}$.
Path $2$: Starts at $70 \text{ V}$,ends at $60 \text{ V}$. $|\Delta V_2| = |60 - 70| = 10 \text{ V}$.
Path $3$: Starts at $80 \text{ V}$,ends at $60 \text{ V}$. $|\Delta V_3| = |60 - 80| = 20 \text{ V}$.
Path $4$: Starts at $70 \text{ V}$,ends at $50 \text{ V}$. $|\Delta V_4| = |50 - 70| = 20 \text{ V}$.
Comparing the magnitudes: $|\Delta V_1| = 40 \text{ V}$,$|\Delta V_3| = 20 \text{ V}$,$|\Delta V_4| = 20 \text{ V}$,$|\Delta V_2| = 10 \text{ V}$.
Therefore,the order of magnitude of work done is $1 > 3 = 4 > 2$.

(C) The electric potential $V$ at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.
Since points $A, B, C, D$ and $E$ all lie on the circumference of a circle centered at the point charge $q$,they are all at the same distance $r$ from the charge.
Therefore,all these points are at the same electric potential,making the path $ABCDE$ an equipotential surface.
By definition,the work done $W$ in moving a charge $q_0$ between two points at the same potential is $W = q_0(V_f - V_i) = q_0(V - V) = 0$.
Thus,the work done in moving a charge from point $A$ to any of the points $B, C, D$ or $E$ is zero.

(B) The electric potential $V$ at any point on a circular path of radius $r$ due to a charge $q_{2}$ at the center is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r}$.
Since $q_{2}$ and $r$ are constant,the potential $V$ is constant at all points on the circular path,making it an equipotential surface.
The work done $W$ in moving a charge $q_{1}$ between two points $A$ and $B$ is given by $W = q_{1}(V_{B} - V_{A})$.
Since the path is equipotential,the potential at the starting point and the ending point (after one full rotation) is the same,i.e.,$V_{A} = V_{B}$.
Therefore,the work done $W = q_{1}(V - V) = 0$.

(C) The electric potential $V$ at any point on the circumference of a circle of radius $R$ due to a charge $+q$ at its centre is given by $V = \frac{kq}{R}$.
Since the potential is the same at all points on the circumference,the circle represents an equipotential surface.
The work done $W$ in moving a charge $q_0$ between two points $B$ and $C$ on an equipotential surface is given by $W = q_0(V_C - V_B)$.
Since $V_B = V_C$,the work done $W = q_0(0) = 0$.

(C) The Assertion is correct because if two equipotential surfaces were to intersect,the point of intersection would have two different values of electric potential,which is physically impossible.
The Reason is incorrect because equipotential surfaces are not necessarily parallel to each other. For example,for a point charge,the equipotential surfaces are concentric spheres,and for a line charge,they are coaxial cylinders. They can have any shape depending on the charge distribution.

(N/A) The system consists of two equal and opposite charges separated by a distance of $6 \; cm$.
An equipotential surface is defined as a surface where the electric potential is the same at every point.
For a system of two equal and opposite charges (an electric dipole),the plane that perpendicularly bisects the line joining the two charges is an equipotential surface with a potential of $0 \; V$.
This plane is located at the mid-point of the line $AB$,i.e.,at a distance of $3 \; cm$ from each charge.
$(b)$ The electric field lines always originate from the positive charge and terminate at the negative charge.
At any point on the equipotential surface (the perpendicular bisector plane),the electric field vector is directed from the positive charge towards the negative charge.
Therefore,the direction of the electric field at every point on this surface is normal to the plane and directed along the line $AB$ (from $A$ to $B$).

(N/A) Equidistant planes parallel to the $x-y$ plane are the equipotential surfaces.
$(b)$ Planes parallel to the $x-y$ plane are the equipotential surfaces,but the spacing between them decreases as the field magnitude increases.
$(c)$ Concentric spheres centered at the origin are the equipotential surfaces.
$(d)$ Near the grid,the equipotential surfaces have a periodically varying shape. At large distances from the grid,these surfaces gradually become planes parallel to the grid.

(N/A) An equipotential surface is an imaginary surface in an electric field where the electric potential is the same at every point.
$(1)$ For a single point charge $q$,the potential at a distance $r$ is $V = \frac{kq}{r}$. Since $V$ is constant for a constant $r$,the equipotential surfaces are concentric spheres centered at the charge.
$(2)$ For a dipole ($+q$ and $-q$),the potential is zero at the mid-plane. The equipotential surfaces are distorted spheres,closer together near the charges and further apart in the middle.
$(3)$ For two like charges ($+q$ and $+q$),the potential is high near the charges. The equipotential surfaces are distorted spheres that do not intersect the mid-plane.
$(4)$ For a uniform electric field,the equipotential surfaces are a set of parallel planes perpendicular to the direction of the electric field lines.