JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 25)
The electric field in an electromagnetic wave is given by E = (50 NC$$-$$1) sin$$\omega$$ (t $$-$$ x/c)
The energy contained in a cylinder of volume V is 5.5 $$\times$$ 10$$-$$12 J. The value of V is _____________ cm3. (given $$\in$$0 = 8.8 $$\times$$ 10$$-$$12C2N$$-$$1m$$-$$2)
The energy contained in a cylinder of volume V is 5.5 $$\times$$ 10$$-$$12 J. The value of V is _____________ cm3. (given $$\in$$0 = 8.8 $$\times$$ 10$$-$$12C2N$$-$$1m$$-$$2)
Answer
500
Explanation
$$E = 50\sin \left( {\omega t - {\omega \over c}.\,x} \right)$$
Energy density = $${1 \over 2}{ \in _0}E_0^2$$
Energy of volume $$V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$$
$${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$$
$$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$$
= .0005 $$\times$$ 106 (c.m)3
= 500 (c.m)3
Energy density = $${1 \over 2}{ \in _0}E_0^2$$
Energy of volume $$V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$$
$${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$$
$$V = {{5.5 \times 2} \over {2500 \times 8.8}} = .0005{m^3}$$
= .0005 $$\times$$ 106 (c.m)3
= 500 (c.m)3
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