JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 11)
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 : $${1 \over {4\,\pi { \in _0}}} = $$ 9 $$ \times $$ 109 N $$-$$ m2/C2 ]
[ Take : $${1 \over {4\,\pi { \in _0}}} = $$ 9 $$ \times $$ 109 N $$-$$ m2/C2 ]
20 mm
32 mm
28 mm
16 mm
Explanation
Energy of sphere = $${{{Q^2}} \over {2C}}$$
$$\therefore\,\,\,$$ $${{16 \times {{10}^{ - 12}}} \over {2C}}$$ = 4.5
$$ \Rightarrow $$$$\,\,\,$$ C = $${{16 \times {{10}^{ - 12}}} \over 9}$$
We know capacity of spherical conductor,
C = 4$$\pi $$$$\varepsilon $$0R
$$\therefore\,\,\,$$ 4$$\pi $$$$\varepsilon $$0R = $${{16 \times {{10}^{ - 12}}} \over 9}$$
$$ \Rightarrow $$$$\,\,\,$$ R = $${1 \over {4\pi {\varepsilon _0}}} \times {{16 \times {{10}^{ - 12}}} \over 9}$$
= 9 $$ \times $$ 109 $$ \times $$ $${{16 \times {{10}^{ - 12}}} \over 9}$$
= 16 mm
$$\therefore\,\,\,$$ $${{16 \times {{10}^{ - 12}}} \over {2C}}$$ = 4.5
$$ \Rightarrow $$$$\,\,\,$$ C = $${{16 \times {{10}^{ - 12}}} \over 9}$$
We know capacity of spherical conductor,
C = 4$$\pi $$$$\varepsilon $$0R
$$\therefore\,\,\,$$ 4$$\pi $$$$\varepsilon $$0R = $${{16 \times {{10}^{ - 12}}} \over 9}$$
$$ \Rightarrow $$$$\,\,\,$$ R = $${1 \over {4\pi {\varepsilon _0}}} \times {{16 \times {{10}^{ - 12}}} \over 9}$$
= 9 $$ \times $$ 109 $$ \times $$ $${{16 \times {{10}^{ - 12}}} \over 9}$$
= 16 mm
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