The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 μ C charge, its radius will be: [ Take: (1)/(4πε₀) = 9 × 109 N - m2/C2 ] (a) 20 mm (b) 32 mm (c) 28 mm (d) 16 mm

Answer: (d). U = Q²/2C = (Q²)/(8πε₀ R) ⇒ R = (1)/(4πε₀)²/2U R = 9×10⁹×(4×10⁻⁶)²2×4.5 = (0.144)/(9) R = 0.016 m = 16 mm So the answer is (d) 16 mm.

The energy stored in the electric field produced by a metal sphere is… | JEE Main 2017 (8 Apr, Shift 1)

Q13JEE Main 2017 · 8 APR · SHIFT 1Medium

Capacitance of a Conductor

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: $\frac{1}{4\pi\epsilon_0} =$ 9 $\times$ 109 N $\text{-}$ m2/C2 ]

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Answer: (d)

$U = \frac{Q^2}{2C} = \frac{Q^2}{8\pi\varepsilon_0 R} \Rightarrow R = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q^2}{2U}$
$R = \frac{9\times10^{9}\times(4\times10^{-6})^2}{2\times4.5} = \frac{0.144}{9}$
$R = 0.016\text{ m} = 16\text{ mm}$
So the answer is (d) 16 mm.