Question

Three concentric conducting spheres of radii $R$, $2R$ and $4R$. The inner and outer spheres are connected with a conducting wire while middle one is given a charge $Q$, then.

• A. The charge on inner sphere is $-\frac{Q}{4}$
• B. The charge on outer sphere is $\frac{Q}{3}$
• C. Potential of inner sphere is $\frac{Q}{16\pi\epsilon_0 R}$
• D. Potential of middle sphere is $\frac{5Q}{48\pi\epsilon_0 R}$

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The problem involves three concentric conducting spheres with radii $R$, $2R$, and $4R$. Let's denote them as Sphere 1, Sphere 2, and Sphere 3, respectively.

• Sphere 1 (radius $R$) has charge $Q_1$.
• Sphere 2 (radius $2R$) is given a charge $Q$.
• Sphere 3 (radius $4R$) has charge $Q_3$.

The inner and outer spheres (Sphere 1 and Sphere 3) are connected by a conducting wire. This implies that they are at the same electric potential. Let this common potential be $V$.

We use the formula for the potential of concentric conducting spheres. The potential at the surface of a sphere is the sum of potentials due to all charges:

$V = k \sum \frac{Q_{inside}}{R_{current}} + k \sum \frac{Q_{outside}}{R_{outside}}$

where $k = \frac{1}{4\pi\epsilon_0}$.

1. Potential of the inner sphere ($V_1$) at radius $R$:

The potential at the surface of Sphere 1 is due to $Q_1$ on Sphere 1, $Q$ on Sphere 2 (which is outside Sphere 1), and $Q_3$ on Sphere 3 (which is outside Sphere 1).

$V_1 = k \frac{Q_1}{R} + k \frac{Q}{2R} + k \frac{Q_3}{4R}$

2. Potential of the outer sphere ($V_3$) at radius $4R$:

The potential at the surface of Sphere 3 is due to $Q_1$ on Sphere 1 (which is inside Sphere 3), $Q$ on Sphere 2 (which is inside Sphere 3), and $Q_3$ on Sphere 3.

$V_3 = k \frac{Q_1}{4R} + k \frac{Q}{4R} + k \frac{Q_3}{4R}$

3. Equating potentials ($V_1 = V_3$):

Since Sphere 1 and Sphere 3 are connected, their potentials are equal:

$k \left( \frac{Q_1}{R} + \frac{Q}{2R} + \frac{Q_3}{4R} \right) = k \left( \frac{Q_1}{4R} + \frac{Q}{4R} + \frac{Q_3}{4R} \right)$

Multiply by $\frac{4R}{k}$:

$4Q_1 + 2Q + Q_3 = Q_1 + Q + Q_3$

$4Q_1 + 2Q = Q_1 + Q$

$3Q_1 = -Q$

$Q_1 = -\frac{Q}{3}$

4. Total charge on the connected system:

A crucial assumption in such problems, if not explicitly stated otherwise, is that the connected conductors (Sphere 1 and Sphere 3 in this case) were initially uncharged, or that they form an isolated system with no net charge initially. Therefore, the total charge on the combined system of Sphere 1 and Sphere 3 must be zero.

$Q_1 + Q_3 = 0$

Using $Q_1 = -\frac{Q}{3}$:

$-\frac{Q}{3} + Q_3 = 0$

$Q_3 = \frac{Q}{3}$

Now we can evaluate the given options:

A. The charge on inner sphere is $-\frac{Q}{4}$

Our calculated charge on the inner sphere is $Q_1 = -\frac{Q}{3}$.

So, option A is incorrect.

B. The charge on outer sphere is $\frac{Q}{3}$

Our calculated charge on the outer sphere is $Q_3 = \frac{Q}{3}$.

So, option B is correct.

C. Potential of inner sphere is $\frac{Q}{16\pi\epsilon_0 R}$

We need to calculate $V_1$ using $Q_1 = -\frac{Q}{3}$ and $Q_3 = \frac{Q}{3}$.

$V_1 = k \left( \frac{Q_1}{R} + \frac{Q}{2R} + \frac{Q_3}{4R} \right)$

$V_1 = k \left( \frac{-Q/3}{R} + \frac{Q}{2R} + \frac{Q/3}{4R} \right)$

$V_1 = \frac{kQ}{R} \left( -\frac{1}{3} + \frac{1}{2} + \frac{1}{12} \right)$

$V_1 = \frac{kQ}{R} \left( \frac{-4 + 6 + 1}{12} \right)$

$V_1 = \frac{kQ}{R} \left( \frac{3}{12} \right) = \frac{kQ}{R} \left( \frac{1}{4} \right)$

Substituting $k = \frac{1}{4\pi\epsilon_0}$:

$V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q}{4R} = \frac{Q}{16\pi\epsilon_0 R}$

So, option C is correct.

D. Potential of middle sphere is $\frac{5Q}{48\pi\epsilon_0 R}$

The potential of the middle sphere ($V_2$) at radius $2R$ is due to $Q_1$ on Sphere 1 (inside Sphere 2), $Q$ on Sphere 2, and $Q_3$ on Sphere 3 (outside Sphere 2).

$V_2 = k \left( \frac{Q_1}{2R} + \frac{Q}{2R} + \frac{Q_3}{4R} \right)$

Substitute $Q_1 = -\frac{Q}{3}$ and $Q_3 = \frac{Q}{3}$:

$V_2 = k \left( \frac{-Q/3}{2R} + \frac{Q}{2R} + \frac{Q/3}{4R} \right)$

$V_2 = \frac{kQ}{R} \left( -\frac{1}{6} + \frac{1}{2} + \frac{1}{12} \right)$

$V_2 = \frac{kQ}{R} \left( \frac{-2 + 6 + 1}{12} \right)$

$V_2 = \frac{kQ}{R} \left( \frac{5}{12} \right)$

Substituting $k = \frac{1}{4\pi\epsilon_0}$:

$V_2 = \frac{1}{4\pi\epsilon_0} \frac{5Q}{12R} = \frac{5Q}{48\pi\epsilon_0 R}$

So, option D is correct.

Options B, C, and D are correct.

Explanation of the solution:

1. Define charges $Q_1, Q, Q_3$ for the inner, middle, and outer spheres, respectively.
2. Write expressions for the potential of the inner sphere ($V_1$) and outer sphere ($V_3$) using the superposition principle for concentric shells.
3. Equate $V_1$ and $V_3$ because the inner and outer spheres are connected by a conducting wire. This yields $Q_1 = -Q/3$.
4. Assume the total charge on the connected system (inner and outer spheres) is zero, as is standard for ungrounded connected conductors without initial charge specified. This gives $Q_1 + Q_3 = 0$, leading to $Q_3 = Q/3$.
5. Calculate the potential of the inner sphere ($V_1$) and the middle sphere ($V_2$) using the determined charges $Q_1 = -Q/3$ and $Q_3 = Q/3$.
6. Compare the calculated values with the given options.