Consider two thin, concentric spherical conductors of radii R1 and R2 where R2 < R1. The conductors are both initially uncharged. If a charge +Q is placed on the outer conductor (of radius R1) the potential on the inner conductor is

Answer 1

To find the potential on the inner conductor (of radius \( R_2 \)) when a charge \( +Q \) is placed on the outer conductor (of radius \( R_1 \)), we can use the principles of electrostatics for spherical conductors:

1. Charge Distribution: Since the outer conductor has a charge \( +Q \), the inner conductor will respond by inducing a charge on its surface. Because the two conductors are concentric and electrically isolated, the inner conductor does not have any net charge (since it was initially uncharged). This means that there will be a negative charge \( -Q \) induced on the inner surface of the outer conductor. Consequently, the induced charge \( -Q \) on the inner conductor's surface remains. Also, the outer surface of the outer conductor will have a charge of \( +Q \) as it is neutral and the total charge is conserved.

2. Electric Field: The electric field in the region between the two conductors (i.e., for \( R_2 < r < R_1 \)), is due to the charge \( +Q \) on the outer conductor. The electric field \( E \) at a distance \( r \) from the center (where \( R_2 < r < R_1 \)) is given by Gauss's law:

   \[ E(r) = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} \]

   Here, \( \epsilon_0 \) is the permittivity of free space.

3. Potential Calculation: The potential difference \( V \) between two points in an electric field can be calculated by integrating the electric field. The potential \( V \) at radius \( R_2 \) can be found by calculating the work done in bringing a unit positive charge from a reference point (we can take infinity for practical purposes) to that point:

   \[ V(R_2) = V(\infty) - V(R_2) = -\int_{\infty}^{R_2} E(r) , dr \]

   Thus,

   \[ V(R_2) = -\int_{\infty}^{R_2} \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r^2} , dr \]

   Calculating this integral:

   \[ V(R_2) = -\frac{1}{4\pi\epsilon_0} \cdot Q \left[ -\frac{1}{r} \right]_{\infty}^{R_2} = -\frac{1}{4\pi\epsilon_0} \cdot Q \left( 0 - \left( -\frac{1}{R_2} \right) \right) = \frac{Q}{4\pi\epsilon_0 R_2} \]

Thus, the potential on the inner conductor (of radius \( R_2 \)) when a charge \( +Q \) is placed on the outer conductor (of radius \( R_1 \)) is:

\[ V(R_2) = \frac{Q}{4\pi\epsilon_0 R_2} \]