Question

A thin charged spherical shell of radius \(10\,cm\) has a uniform surface charge density of \(20\,pC/cm^{2}\). The electric potential at a point \(8\,cm\) from the centre of the shell is:

• A. \(180\pi\,V\)
• B. \(360\pi\,V\)
• C. \(720\pi\,V\)
• D. Zero

Hint:

Inside a uniformly charged spherical shell, \[ \boxed{E=0} \] but \[ \boxed{V=\frac{kQ}{R}} \] which is constant everywhere inside the shell. Do not use \(V=\dfrac{kQ}{r}\) for points inside a spherical shell.

Answer 1

The Correct Option is C

Concept: For a uniformly charged spherical shell, \[ \boxed{V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}} \] at every point inside the shell as well as on its surface. The total charge on the shell is \[ \boxed{Q=\sigma(4\pi R^2)} \] where \[ \sigma=\text{Surface charge density}, \] \[ R=\text{Radius of the shell}. \] Since the observation point is inside the shell (\(r\lt R\)), the potential remains constant throughout the interior.

Step 1: Write the given data.
Given, \[ R=10\,cm=0.1\,m, \] \[ \sigma=20\,pC/cm^2. \] Convert the surface charge density into SI units. Since \[ 1\,pC=10^{-12}C, \] and \[ 1\,cm^2=10^{-4}m^2, \] therefore, \[ \sigma = 20\times10^{-12}\times10^{4} = 2\times10^{-7}C/m^2. \]

Step 2: Calculate the total charge on the shell.
Using \[ Q=\sigma(4\pi R^2), \] we obtain \[ Q = 2\times10^{-7}\times4\pi\times(0.1)^2. \] Since \[ (0.1)^2=0.01, \] \[ Q = 2\times10^{-7}\times4\pi\times0.01 = 8\pi\times10^{-9}C. \] Thus, \[ \boxed{Q=8\pi\times10^{-9}C.} \]

Step 3: Calculate the electric potential inside the shell.
Using, \[ V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}, \] where \[ \frac{1}{4\pi\varepsilon_0}=9\times10^9. \] Substituting, \[ V = 9\times10^9 \times \frac{8\pi\times10^{-9}}{0.1}. \] Therefore, \[ V = 720\pi\,V. \] Hence, \[ \boxed{V=720\pi\,V.} \] Thus, the correct answer is \[ \boxed{\textbf{Option (C)}}. \]

Important Observation: Using the standard electrostatic formula, \[ \boxed{V=720\pi\,V.} \] Hence Option (C) is mathematically correct.