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2. Electric Potential and Capacitance
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In a uniform electric field, a cube of side $1\ cm$ is placed. The total energy stored in the cube is $8.85\ μJ$ . The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is
A
$\frac{1}{{5\sqrt 2 }}\ V-m$
B
$100\sqrt 2 \ V-m$
C
$5\sqrt 2 \ V-m$
D
$10\sqrt 2 \ V-m$
Solution
Energy $=\frac{1}{2} \in_{0} \mathrm{E}^{2}$ ( volume )
$8.85 \times 10^{-6}=\frac{1}{2} \times 8.85 \times 10^{-12} \mathrm{E}^{2}\left(10^{-6}\right)$
$\mathrm{E}=\sqrt{2} \times 10^{6} \mathrm{\,V} / \mathrm{m}$
flux $(\phi)=E A$
$=\sqrt{2} \times 10^{+6} \times 10^{-4}=100 \sqrt{2}(\mathrm{V}-\mathrm{m})$
Standard 12
Physics
