Question

A solid conducting sphere of radius \(R\) has a charge \(Q\) evenly distributed over its surface, producing an electric potential \(V_{0}\) at the surface. How much charge must be added to the sphere to increase the potential at the surface to \(2 V_{0} ?\) a) \(Q / 2\) b) \(Q\) c) \(2 Q\) d) \(Q^{2}\) e) \(2 Q^{2}\)

Short answer

Answer: (b) Q

Step-by-step solution

Write down the expression of the potential at the surface of the sphere

The potential (V) at the surface of a conducting sphere with charge Q and radius R, is given by the formula: V = \frac{kQ}{R} Here, k is the electrostatic constant.

Calculate the initial potential V_0

The initial potential in the problem, V_0, is given by replacing V by V_0: V_0 = \frac{kQ}{R}

Write down the expression for the doubled potential

Now, double the potential, which we denote as 2V_0. This is equal to the potential of the charged sphere with the new charge, say Q' (which is the sum of the initial charge Q and the added charge x) and the same radius R. We express this as: 2V_0 = \frac{kQ'}{R}

Substitute the expression for V_0 in the above equation

Now, we can substitute the expression for V_0 from step 2 into the equation in step 3: 2(\frac{kQ}{R}) = \frac{kQ'}{R}

Solve for the new charge, Q'

The added charge x is equal to the difference between the new charge Q' and the initial charge Q. We need to solve the equation in step 4 for Q' and then find the value of x: 2(\frac{kQ}{R}) = \frac{k(Q + x)}{R} Divide both sides by k and multiply both sides by R: 2Q = Q + x

Calculate the added charge x

Isolate x to find its value: x = 2Q - Q x = Q Hence, the required added charge x to double the surface potential is equal to the initial charge Q. The answer is (b) Q.