Question

What is the potential at the centre of a spherical shell ofradius Rthat has a charge Quniformly distributed over its surface? Don’t do an integral of the electric field; just add up the contributions to the potential by the charges.

Short answer

The potential at the centre of a spherical shell is, $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• Charge, Q
• Radius, R
• Small charges, dQ
• Potential energy, V

Understanding Electric Potential:

Electric potential is defined as the effort required to shift a unit charge from one point to another while battling the effects of an electric field. The energy gained by an object when it is moved against an electric field is known as the electric potential energy. Divide the potential energy by the amount of charge to get the charge's electric potential.

Determination of the potential at the centre of a spherical shell.

In this problem we calculate the electric potential in the centre of a spherical shell of radiusRand chargeQ. Since the electric field inside the shell is zero, the potential at the centre is the same as the potential on the surface of the sphere. We can consider the sphere as a sum of infinitely many small chargesdQthat are at a distanceRfrom the centre. Each charge generates a potential

$dV=\frac{1}{4\pi {\epsilon }_0}\frac{dQ}{R}$

so, the total potential is

$$\begin{gathered} V=\sum dv \\ V=\sum \frac{1}{4\pi {\epsilon }_0}\frac{dQ}{R} \\ V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R} \end{gathered}$$