JEE Main On 8 April 2017 Question 9
Question: The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains $ 4\mu C $ charge, its radius will be: [Take: $ \frac{1}{4\pi {\varepsilon_0}}=9\times 10^{9}N-m^{2}/C^{2} $ ] [JEE Online 08-04-2017]
Options:
A) 32 mm
B) 16 mm
C) 28 mm
D) 20 mm
Show Answer
Answer:
Correct Answer: B
Solution:
- Energy of sphere $ =\frac{Q^{2}}{2C} $ $ 4.5=\frac{16\times {10^{-12}}}{2C} $ $ C=\frac{16\times {10^{-12}}}{9}=4\pi {\varepsilon_0}R $ $ R=\frac{16\times {10^{-12}}}{9}\times \frac{1}{4\pi {\varepsilon_0}} $ $ =9\times 10^{9}\times \frac{16}{9}\times {10^{-12}} $ $ =16\times {10^{-3}}=16mm $
