Question

Consider two concentric conducting spheres of radii $R$ and $2R$ respectively. The inner sphere is given a charge $+Q$. The other sphere is grounded. The potential at $r = \frac{3R}{2}$ is

• A. 1/4π∈0 Q/6R
• B. 0
• C. 1/4π∈0 2Q/3R
• D. 1/4π∈0 Q/R

Answer 1

The Correct Option is A

Given: Two concentric conducting spheres with radii \( R \) and \( 2R \). The inner sphere is given a charge \( +Q \), and the outer sphere is grounded (i.e., its potential is zero). We are to find the potential at a point where \( r = \frac{3R}{2} \), which lies between the two spheres.

Concept: When the outer sphere is grounded, it induces a charge \( -Q \) on its inner surface (to neutralize the electric field outside), and a charge \( +Q \) appears on its outer surface to maintain neutrality. The potential at any point between the spheres (i.e., for \( R < r < 2R \)) is due to both the inner sphere and the induced charge on the inner surface of the outer sphere. 

Step-by-step calculation: Potential at a point \( r \) due to a spherical shell of charge \( Q \) is: \[ V(r) = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q_{\text{enclosed}}}{r} \] Let us consider: - Inner sphere at radius \( R \) has charge \( +Q \) - Induced charge \( -Q \) appears on the inner surface of the outer sphere at radius \( 2R \) So at \( r = \frac{3R}{2} \), total potential is the sum of potentials due to: 1. Charge \( +Q \) at center (inner sphere): \[ V_1 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{3R/2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Q}{3R} \] 2. Charge \( -Q \) at radius \( 2R \): Since the field inside a shell is same as if all the charge were at center, potential due to shell at point inside is: \[ V_2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{-Q}{2R} \] 

Total Potential: \[ V = V_1 + V_2 = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{3R} - \frac{Q}{2R} \right) = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(4Q - 3Q)}{6R} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{6R} \] 

Final Answer: \( \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{6R} \)