Question

A4 mu parallel plate capacitor is connected to a 5 volt battery the work done by an external agent is slowly increase the separation between the plates to double its original value is


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Answer 1

Answer:

the answer is 20uj

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Answer 2

Answer:

The required work done = 2mJ

Explanation:

• Initially the energy that is stored in the capacitor is = 1/2 C*V^2 or

        Q^2/ 2*C

• Now when the separation between the plates gets doubled, then the charge remains constant and their capacitance gets halved.
• therefore the work done in separation of the plates is basically the difference in stored energy, which is : (Q^2 / 2*0.5*C) - (Q^2 / 2*C)
• It represents that if we double the initial energy then less will be the initial energy
• Therefore the work done = Q^2 / 2*C

so on calculating :  (2*10–4 )^2 / ( 2* 10 *10^-6 ) we get, the required work done is 2mJ

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