Math, asked by teamahardrick, 1 month ago

PLease help me it is really important ( no links ) and please explain why your answer is correct..

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Answers

Answered by yendavasudhagmailcom
4

Answer:

The energy initially stored by the capacitor is 1/2 C*V^2 or Q^2 / 2*C

When the plate separation is doubled, the charge remains constant and the capacitance is halved. The work done in separating the plates is the difference in energy stored: this is

Q^2 / 2*0.5*C - Q^2 / 2*C This represents double the initial energy less the initial energy. The work done in adding this energy is Q^2 / 2*C

Assigning values: ( 2*10–4 )^2 / ( 2* 10 *10^-6 ) = 2 mJ

Step-by-step explanation:

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