A parallel-plate capacitor has fixed charges + Q and Q . The separation of the p

Question

A parallel-plate capacitor has fixed charges +Q and Q. The separation of the plates is then halved.

Problem 17.56 A parallel-plate capacitor has fixed charges +Q and -Q The separation of the plates is then halved. Part A By what factor does the energy stored in the electric field change? Express your answer using two significant figures. PEfinal PEinitial Submit My Answers Give U incorrect; Try Again; 5 attempts remaining The correct answer does not depend on: U Part B How much work must be done to reduce the plate separation from d to 1 d? The area of each plate is A. Express your answer in terms of the variables 2, d, A and the constant Eo APE Submit My Answers Give up Incorrect Try Again: 5 attempts remaining

Explanation / Answer

Capacitance of a capacitor is given by the formula

C=A/d

And E = Q^2/2C

Combining above both
E=Q^2/2C=dQ^2/2A
Now, if d is halved then E is also halved

Part A) energy stored PE(final)/PE(initial) = 1/2

Part B) since we know ,
work done is change in energy stored

Now , W=E2-E1
=(d2-d1)Q^2/2A

Since d2 = d/2 and d1 = d

Substituting in above formula

Work done = (d/2 - d) Q^2/2A

= -1/2*d*( Q^2/2A)

Therefore PE = -1/2*d*( Q^2/2A)

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