Q.In a region of constant potential

The problem asks for the net electrostatic energy of the two charges. This is not just the energy stored in the field between them — it also includes the energy each charge has because it sits in the external electric field given. The total electrostatic energy of a system of point charges in an external field is:

The first term is the potential energy of each charge in the external potential $V_{\text{ext}}$ (which comes from the given field). The second term is the mutual electrostatic potential energy between the two charges themselves — the familiar $k q_1 q_2 / r$.

We are given the external electric field as $E = 1.8 \times 10^5 \, r^{-2} \; \text{N/C}$, where $r$ is the distance from the origin. This field is radial and points outward (since it's positive). The potential corresponding to a radial field $E = \frac{C}{r^2}$ is $V = -\int E \, dr = \frac{C}{r}$, taking the zero of potential at infinity. So the external potential at a distance $r$ from the origin is:

Now let’s compute step by step.

1. Convert distances to metres. The charges are at $x = -2.7$ cm and $x = +2.7$ cm.

Both are the same distance from the origin.

2. Compute the external potential energy for each charge. For $q_1 = +6\ \mu\text{C} = 6 \times 10^{-6}\ \text{C}$ at $r = 0.027$ m:

For $q_2 = -3\ \mu\text{C} = -3 \times 10^{-6}\ \text{C}$ at the same distance:

So the total external contribution is $40 + (-20) = 20$ J.

3. Compute the mutual electrostatic energy between the two charges. The distance between them is $2.7 - (-2.7) = 5.4$ cm $= 0.054$ m. Using $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$: $$$$

Concept. When a resistor bridges two nodes of a network, the current through it is set by the potential difference across it: $I = (V_A - V_B)/6$. So the task is to find $V_A$ and $V_B$ from the rest of the circuit and then divide by $6\,\Omega$.

Method.

1. Treat the outer resistors as the arms of a bridge feeding nodes A and B from the supply. …

For the bridge/network shown, apply Kirchhoff's laws to the loops meeting at junctions A and B. Assigning branch currents and writing the loop (KVL) and junction (KCL) equations for the mesh containing the $6~\Omega$ arm and reducing the resulting simultaneous equations gives the current in that arm as

$I_{6\Omega}=0.25~\text{A}.$ …

The key idea is that the electric field due to a point charge is conservative. That means the work done to move a test charge between two points is path-independent — it depends only on the electrostatic potential at the start and end points. So instead of integrating force along a path, we can simply compute the potential difference and multiply by the test charge.

For a point charge $q$, the potential at a distance $r$ is $V = \frac{kq}{r}$, where $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$. The work done by an external agent in moving a charge $q_0$ from A to B (without acceleration) equals the change in potential energy: $W = q_0 (V_B - V_A)$.

Let’s go step by step.

1. Identify the given data

   Source charge: $q = +20\ \mu\text{C} = 20 \times 10^{-6}\ \text{C}$

   Test charge: $q_0 = +2\ \mu\text{C} = 2 \times 10^{-6}\ \text{C}$

   Distances: $r_A = 4\ \text{cm} = 0.04\ \text{m}$, $r_B = 5\ \text{cm} = 0.05\ \text{m}$

   Constant: $k = 9 \times 10^9\ \text{N m}^2/\text{C}^2$

2. Compute the potentials at A and B

   At A:

At B:

3. Find the potential difference

The negative sign means potential is lower at B (farther from the positive source). The work done by an external agent to move the positive test charge from A to B is against the field, so we use the magnitude of the change.

4. Calculate the work done $W = q_0 (V_B - V_A) = (2 \times 10^{-6})(-0.9 \times 10^6) = -1.8\ \text{J}$ …

The key idea is that the electric field $\vec{E}$ is related to the potential $V$ by $\vec{E} = -\nabla V$. This means we take partial derivatives of $V$ with respect to each coordinate, then flip the sign. The gradient points in the direction of steepest increase of $V$; the field points opposite, toward decreasing potential.

Let’s work through it step by step.

1. Recall the definition. In Cartesian coordinates, the electric field components are

Here $V$ depends only on $x$ and $y$, so $E_z = 0$.

2. Compute the partial derivative with respect to $x$. Given $V = \frac12(y^2 - 4x) = \frac12 y^2 - 2x$,

Therefore,

3. Compute the partial derivative with respect to $y$.

Therefore,

At $y = 1$, this gives $E_y = -1$.

4. Assemble the vector. So $\vec{E} = 2\hat{i} + (-1)\hat{j} = 2\hat{i} - \hat{j}\ \text{V/m}$. Wait — that’s not among the options exactly? Let’s check carefully: Option (A) is $2\hat{i} + \hat{j}$, option (C) is $2\hat{i} - \hat{j}$. Did we get the sign wrong on $E_y$? Let’s re-evaluate: $$$$

Concept and Intuition

Electric potential is a scalar quantity that adds algebraically, while electric field is a vector that requires directional consideration. The key insight is that we first use the condition of zero potential to locate point P, then calculate the electric field at that location.

For a point charge $q$, the potential at distance $r$ is $V = \frac{kq}{r}$ and the electric field magnitude is $E = \frac{k|q|}{r^2}$, where $k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2$.

Finding Point P

Let's place the $+2\mu C$ charge at the origin and the $-4\mu C$ charge at distance $3\,\text{m}$. Let point P be at distance $x$ from the positive charge.

1. Set up the potential equation

The total potential at P is the algebraic sum of potentials from both charges:

2. Solve for the position x

Dividing by $k \times 10^{-6}$:

Cross-multiplying:

So point P is $1\,\text{m}$ from the $+2\mu C$ charge and $2\,\text{m}$ from the $-4\mu C$ charge.

Calculating the Electric Field at P

3. Find the electric field from each charge

The electric field from the $+2\mu C$ charge at distance $1\,\text{m}$:

$E_1 = \frac{k \times 2 \times 10^{-6}}{1^2} = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{1} = 18{,}000\,\text{N/C}$ …

The key idea is that the potential at the centre of a uniformly charged ring is easy to compute — every point on the ring is at the same distance from the centre, so the potential is just $\frac{1}{4\pi\varepsilon_0}\frac{q}{R}$. For a point on the axis, the distance to each ring element is $\sqrt{R^2 + x^2}$, so the potential is $\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{R^2 + x^2}}$.

Here we have two rings on the same axis, separated by $\sqrt{3}R$. One has charge $+q$, the other $-q$. We want the potential difference between their centres. That means: potential at centre of ring 1 minus potential at centre of ring 2. Each centre feels the potential from its own ring (easy) plus the potential from the other ring (a little less easy, but still straightforward).

Let’s set it up clearly.

1. Place the rings on the z-axis. Let ring 1 (charge $+q$) lie in the plane $z = 0$, with its centre at $z = 0$. Let ring 2 (charge $-q$) lie in the plane $z = d$, where $d = \sqrt{3}R$, with its centre at $z = d$.

2. Potential at centre of ring 1 ($V_1$).

   • From its own ring: every point on ring 1 is at distance $R$ from its centre, so

• From the other ring (ring 2, charge $-q$): the centre of ring 1 is on the axis of ring 2, at a distance $d = \sqrt{3}R$ from the plane of ring 2. So each point on ring 2 is at distance $\sqrt{R^2 + d^2} = \sqrt{R^2 + 3R^2} = 2R$ from the centre of ring 1. Hence

• Total:

3. Potential at centre of ring 2 ($V_2$).
   • From its own ring (charge $-q$):

• From the other ring (ring 1, charge $+q$): the centre of ring 2 is at distance $d = \sqrt{3}R$ from ring 1, so again the distance is $2R$. Thus …

Concept and Intuition

When two conducting spheres are brought into contact and given a total charge, the charge does not necessarily distribute equally. Instead, charge will flow between the spheres until their electric potentials become equal. This is a fundamental property of conductors in electrostatic equilibrium.

1. Equal Potential: If two conductors are in contact, they form a single equipotential surface. Therefore, the electric potential on the surface of the first sphere ($V_1$) must be equal to the electric potential on the surface of the second sphere ($V_2$).
2. Potential of a Sphere: The electric potential on the surface of an isolated conducting sphere of radius $R$ carrying a charge $Q$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$.
3. Charge Distribution: Since $V_1 = V_2$, it follows that $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$, or $\frac{Q_1}{Q_2} = \frac{R_1}{R_2}$. This means the charge on each sphere is directly proportional to its radius.
4. Conservation of Charge: The total charge given to the system, $Q$, is conserved. So, $Q_1 + Q_2 = Q$.
5. Potential at a Distance: Once the spheres are separated, each acts as an isolated charged sphere. The electric potential at a distance $r$ from the center of a sphere carrying charge $Q$ (where $r$ is greater than the sphere's radius) is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$. This is the same formula used for the potential due to a point charge.

Step-by-Step Solution

1. Identify Given Information and Ratios:

   We are given that the radii of the two metal spheres are in the ratio $4:7$. Let $R_1$ be the radius of the smaller sphere and $R_2$ be the radius of the larger sphere.

   So, $\frac{R_1}{R_2} = \frac{4}{7}$.

   The total charge given to the system is $Q = 8.8 \times 10^{-7}$ C.

   We need to find the potential due to the smaller sphere at a distance of $60$ m from its center.

2. Apply the Equal Potential Condition:

   When the two spheres are in contact, charge redistributes until their potentials are equal. Let $Q_1$ be the charge on the smaller sphere (radius $R_1$) and $Q_2$ be the charge on the larger sphere (radius $R_2$).

   The potential on the surface of each sphere is:

   $V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_1}$

   $V_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2}{R_2}$

   Since $V_1 = V_2$:

   $\frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_1} = \frac{1}{4\pi\epsilon_0} \frac{Q_2}{R_2}$

   This simplifies to:

   $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$

   From this, we get the ratio of charges:

   $\frac{Q_1}{Q_2} = \frac{R_1}{R_2} = \frac{4}{7}$

   So, $Q_1 = \frac{4}{7} Q_2$.

3. Apply Conservation of Charge:

   The total charge given to the system is $Q = Q_1 + Q_2$. …

Concept & Intuition

When a battery is connected between opposite corners A and C of a square loop made of uniform wire, the current splits into two parallel paths: one going through two sides (A→B→C) and the other through the other two sides (A→D→C). Because the wire is uniform, the resistance of each path is the same, so the current divides equally. At the centre of the square, each straight segment of wire produces a magnetic field whose direction is given by the right-hand rule. The fields from opposite sides of the square point in opposite directions and have equal magnitude, leading to cancellation. The net field is therefore zero.

Step-by-step reasoning

1. Current distribution

   The battery is connected between A and C, opposite corners. The square has four equal sides, each of resistance $R$. Path A→B→C has two sides in series, total resistance $2R$. Path A→D→C also has resistance $2R$. Since the two paths are in parallel, the current $I$ from the battery splits equally: each path carries $I/2$.

2. Magnetic field from a single straight wire segment

   For a finite straight wire of length $L$, the magnetic field at a perpendicular distance $d$ from its midpoint is

where $\theta_1$ and $\theta_2$ are the angles from the point to the ends of the wire. For a square of side $a$, the centre is at distance $a/2$ from each side, and each side subtends angles $\theta_1 = \theta_2 = 45^\circ$ at the centre. Hence for one side carrying current $I/2$:

3. Direction of fields from each side Using the right-hand rule:
   • Side AB: current from A to B → field at centre points into the page.
   • Side BC: current from B to C → field at centre points out of the page.
   • Side CD: current from C to D → field at centre points into the page. …

Concept & Intuition

The work done in moving a charge in an electrostatic field is path-independent — it depends only on the starting and ending points. Here, the charge +Q moves along a semicircle from C (midpoint) to D (the point directly opposite C on the other side of the line AB). The fixed charges +q and –q create a potential that is symmetric. The trick: the potential at C is zero (equal distances to +q and –q), but at D the distances are different, so the potential is not zero. The work done by an external agent equals $Q$ times the potential difference $V_D - V_C$. We compute that difference.

Step-by-step solution

1. Set up coordinates and distances

   Place A at $(-L, 0)$ and B at $(L, 0)$. Then C, the midpoint, is at $(0,0)$. The semicircle CRD has center at C and radius $L$ (since AB = 2L, so half is L). Point D is directly opposite C on the semicircle: at $(0, -L)$ if we take the semicircle downward, or $(0, L)$ upward — the exact orientation doesn’t matter; we choose D at $(0, -L)$.

2. Potential at C

   Distance from C to A: $L$. Distance from C to B: $L$.

   Potential at C due to +q at A: $\frac{1}{4\pi\varepsilon_0} \frac{+q}{L}$.

   Potential at C due to –q at B: $\frac{1}{4\pi\varepsilon_0} \frac{-q}{L}$.

   Sum:

3. Potential at D Distance from D $(0,-L)$ to A $(-L,0)$:

Distance from D to B $(L,0)$:

So both distances are equal! That means:

4. Wait — both potentials are zero?

   That would imply zero work. But check the geometry: D is on the perpendicular bisector of AB? Actually, the perpendicular bisector of AB is the vertical line through C. Points on that line are equidistant from A and B. D is at $(0,-L)$, which lies on that line. So indeed D is equidistant from A and B. Therefore $V_D = 0$ as well. So $V_C = V_D = 0$, and the work done $W = Q(V_D - V_C) = 0$. But zero is not among the options — so we must have misinterpreted the semicircle.

5. Reexamine the semicircle CRD

   The semicircle is drawn with diameter CD? Or is C the midpoint of AB and the semicircle goes from C to D, where D is on the line through B? Let’s read: “Charge +q and –q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is…” Usually in such problems, the semicircle has diameter CD, with C at the midpoint and D at the other end of the diameter perpendicular to AB. But here D is likely the point on the opposite side of the circle such that the path goes from C to D along a semicircle of radius L. However, if D is directly above C (on the perpendicular bisector), we got zero. So D must be elsewhere.

6. Correct interpretation: D is on the line AB extended?

   A classic problem: A and B are 2L apart, C is midpoint. A semicircle is drawn with diameter CD where D is on the line AB such that CD = 2L? No — more common: The semicircle is drawn on AB as diameter, so C is the center and the semicircle goes from A to B? But the problem says “along the semicircle CRD” — points C, R, D. So C and D are endpoints of the semicircle. If C is the midpoint of AB, then D must be the point diametrically opposite C on a circle of radius L. That circle’s center is C, so D is at distance L from C. But then D lies on the perpendicular bisector — giving zero. So the only way to get a nonzero answer is if D is not on that bisector.

7. Alternative: The semicircle has diameter AB?

   If the semicircle has AB as diameter, then its center is C, radius L, and the semicircle goes from A to B. But the problem says “semicircle CRD” — so C and D are endpoints, and R is a point on the arc. If C is the midpoint of AB, then D must be the other endpoint of the diameter through C perpendicular to AB. That gives D at $(0, L)$ or $(0, -L)$. Again zero. So something is off.

8. Look at the options: they are nonzero.

   The options involve $\frac{qQ}{2\pi\varepsilon_0 L}$, $\frac{qQ}{3\pi\varepsilon_0 L}$, $-\frac{qQ}{6\pi\varepsilon_0 L}$, $\frac{qQ}{4\pi\varepsilon_0 L}$. The negative sign in (C) hints that work might be negative. Let’s compute the potential at C and at a point D that is on the line AB but on the opposite side of B? For instance, if D is at a distance 2L from A? No.

9. Most plausible standard problem: …

The key concept here is that work done by an external agent in moving a charge in an electrostatic field is path-independent — it depends only on the initial and final positions. For a conservative field, the work done equals the change in potential energy:

where $V$ is the electric potential due to all other charges. So we don’t need to integrate along the semicircle; we just compare potentials at points C and D.

1. Set up the geometry and potentials

   Points A and B are separated by $2L$, so C is the midpoint. Let A be at $x = -L$, B at $x = +L$, and C at $x = 0$. The semicircle CRD has its diameter along AB, so D is directly above C at a distance $L$ (since the semicircle radius is $L$). Coordinates:

   • C = $(0,0)$
   • D = $(0,L)$

2. Potential at C

   Distance from A to C: $L$ (charge $+q$)

   Distance from B to C: $L$ (charge $-q$)

3. Potential at D Distance from A to D: $\sqrt{L^2 + L^2} = L\sqrt{2}$ Distance from B to D: also $L\sqrt{2}$

4. Work done Since $V_C = V_D = 0$, the potential difference is zero: $W = Q \cdot (V_D - V_C) = 0$ …