KCET · Physics · Electrostatics
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 \( V_{1} < V_{2} < V_{3} \)
- C \( V_{1}=V_{2}>V_{3} \)
- D \( V_{1}=V_{2} < V_{3} \)
Answer & Solution
Correct Answer
(C) \( V_{1}=V_{2}>V_{3} \)
Step-by-step Solution
Detailed explanation
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
\(V=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{R} \Rightarrow V \propto \frac{1}{R} \)
\(\Rightarrow V_{3}>V_{1}=V_{2}\)
Thus, electric potential at \( 5 \mathrm{~cm}, 10 \mathrm{~cm} \) and \( 15 \mathrm{~cm} \) are
\(V_{1}=V_{2}>V_{3}\)
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
\(V=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q}{R} \Rightarrow V \propto \frac{1}{R} \)
\(\Rightarrow V_{3}>V_{1}=V_{2}\)
Thus, electric potential at \( 5 \mathrm{~cm}, 10 \mathrm{~cm} \) and \( 15 \mathrm{~cm} \) are
\(V_{1}=V_{2}>V_{3}\)
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