A spherical shell of radius \( 10 \mathrm{~cm} \) is carrying a charge \( \mathrm{q} \). If the electric potential at distances \( 5 \)
\( \mathrm{cm}, 10 \mathrm{~cm} \) and \( 15 \mathrm{~cm} \) from the centre of the spherical shell is \( V_{1}, V_{2} \) and \( V_{3} \) respectively,
then

• A

  \( V_{1}>V_{2}>V_{3} \)
• B

  \( V_{1} < V_{2} < V_{3} \)
• Correct answer
  C

  \( V_{1}=V_{2}>V_{3} \)
• D

  \( V_{1}=V_{2} < V_{3} \)

Solution

It is given,

electric potential at \( 5 \mathrm{~cm}=V_{1} \)
electric potential at \( 10 \mathrm{~cm}=V_{2} \)
electric potential at \( 15 \mathrm{~cm}=V_{3} \)
We know potential inside a conductor is same everywhere, therefore electric potential at \( 5 \mathrm{~cm} \) and \( 10 \mathrm{~cm} \) distance will be
same, that is, \( V_{1}=V_{2} \)
Now, electric potential is given as
\[ \begin{array}{l} V=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{R} \Rightarrow V \propto \frac{1}{R} \\ \Rightarrow V_{3}>V_{1}=V_{2} \end{array} \]
Thus, electric potential at \( 5 \mathrm{~cm}, 10 \mathrm{~cm} \) and \( 15 \mathrm{~cm} \) are
\[ V_{1}=V_{2}>V_{3} \]

Hence, the correct answer is option: 'C'.