Part A An uncharged capacitor is connected to the terminals of a 4.0 V battery,

Question

Part A

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 12 C flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 8.0 V battery, with the positive and negative terminals connected in the same manner as before.

How much additional charge flows to the positive plate?

Part B

Two uncharged metal spheres, spaced 10.0 cmapart, have a capacitance of 28.0 pF .

Part C

A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0 .

Explanation / Answer

A.

initial capacitance will be:

C = Q/V

C = 12*10^-6/4 = 3*10^-6 F

Now capacitance will be same, so new charge will be:

Q1 = C*V1

Q1 = 3*10^-6*8 = 24*10^-6 C= 24 uC

additional charge will be = 24 - 12 = 6 uC

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