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Question

The capacitance of a variable capacitor joined with the battery of $$100$$V is changed from $$2\mu F$$ to $$10\mu F$$. What is the change in the energy stored in it?


A
2×102J
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B
2.5×102
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C
6.5×102J
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D
4×102J
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Solution

The correct option is D $$4\times 10^{-2}$$J
Given:
$$Voltage=V=100V$$
$$C_1=2\mu F$$ (Initial capacitance)
$$C_2=10\mu F$$ (Final capacitance)
As $$E=\dfrac{1}{2}CV^{2}$$
Now
$$E_1=\dfrac{1}{2}C_1V^{2}=\dfrac{1}{2}\times 2\times {10}^{-6}\times V^{2}$$
$$E_2=\dfrac{1}{2}C_2V^{2}=\dfrac{1}{2}\times 10\times {10}^{-6}\times V^{2}$$
$$\Delta E=E_2-E_1=\dfrac{1}{2}\times 10\times {10}^{-6}\times V^{2}-\dfrac{1}{2}\times 2\times {10}^{-6}\times V^{2}$$
$$\Delta E=\dfrac{1}{2}\times 8\times {10}^{-6}\times V^{2}$$
$$\Delta E=\dfrac{1}{2}\times 8\times {10}^{-6}\times {100}^{2}$$
$$\Delta E=4\times 10^{-2}J$$
Hence the correct option is (D).



Physics

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