Problem 71 - A Complete Resource Book in Physics for JEE Main

Question

A conducting hollow sphere of radius \(0.1 \mathrm{~m}\) is given a charge of \(10 \mu \mathrm{C}\). The electric potential on the surface of sphere will be (A) Zero (B) \(3 \times 10^{5} \mathrm{~V}\) (C) \(9 \times 10^{5} \mathrm{~V}\) (D) \(9 \times 10^{9} \mathrm{~V}\)

Step-by-Step Solution

Answer

The electric potential on the surface of the conducting sphere is (C) \(9 \times 10^5\,\mathrm{V}\).

Step 1: Identify the formula for the electric potential of a charged sphere

We will use the formula for the electric potential of a charged sphere, given by: \[V = k\frac{Q}{r}\] where - \(V\) is the electric potential, - \(k\) is the Coulomb constant (\(8.9875 \times 10^9\,\mathrm{Nm^2/C^2}\)), - \(Q\) is the charge of the sphere, and - \(r\) is the radius of the sphere.

Step 2: Plug in the given values

We are given the values for the radius, \(r = 0.1\,\mathrm{m}\), and charge, \(Q = 10 \mu \mathrm{C} = 10 \times 10^{-6}\,\mathrm{C}\). Plugging these values into the formula, we get: \[V = k\frac{Q}{r} = (8.9875 \times 10^9)(\frac{10 \times 10^{-6}}{0.1})\]

Step 3: Calculate the electric potential

Now, we'll simplify the expression for the electric potential: \[V = (8.9875 \times 10^9)(10^5 \times 10^{-6})\] \[V = 8.9875 \times 10^4\]

Step 4: Compare to the given options

Now, let's compare the calculated value of the electric potential to the given options: (A) Zero (B) \(3 \times 10^5\,\mathrm{V}\) (C) \(9 \times 10^5\,\mathrm{V}\) (D) \(9 \times 10^9\,\mathrm{V}\) Since \(8.9875 \times 10^4\) is closest to \(9 \times 10^5\,\mathrm{V}\), the correct answer is (C) \(9 \times 10^5\,\mathrm{V}\).