JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2017
- Hard
- Numerical
The energy stored in the electric field produced by a metal sphere is \(4.5\, J\). lf the sphere contains \(4\,\mu C\) charge, its radius will be.......\(mm\) : [Take : \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N - {m^2}\,/{C^2}\, ]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
Energy of sphere \(=\frac{Q^{2}}{2 C}\)
\(4.5=\frac{16 \times 10^{-12}}{2 C}\)
\(C=\frac{16 \times 10^{-12}}{9}=4 \pi \varepsilon_{0} R\)
(capacity of spherical conductor)
\(\mathrm{R}=\frac{16 \times 10^{-12}}{9} \times \frac{1}{4 \pi \varepsilon_{0}}\)
\(\because \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}\)
\(=9 \times 10^{9} \times \frac{16}{9} \times 10^{-12}=16\, \mathrm{mm}\)
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