Question

1) An isolated capacitor with capacitance C = 4

Answer #1

a) The energy stored in a capacitor is:

U = Q^2/(2C) = (80*10^-6)^2/(2*4*10^-6) = **8 x 10^-4
J**

b) Charge:

The charge on the original capacitor plates is unchanged.

So Q = **80 x 10^-6 C**

c) Capacitance:

Because the charge and charge density on the capacitor plates are
unchanged, the E-field intensity is also unchanged - except where
the conducting material has intruded and "killed" the E-field. So
now the E-field is non-zero only over 1 - 1/3 = 2/3 of its original
distance, so the voltage drop is 2/3 of the orginal. Since V = Q/C,
C = Q/V, and since Q is unchanged but V has changed by factor 2/3,
C has changed by factor 3/2: C' = (3/2)C

So the new capacitance = (3/2) old_capacitance

= (3/2)*4*10^-6

= **6 x 10^-6 F**

d) U' = Q^2/(2C') = (2/3)(Q^2/(2C)) = (2/3)*U = (2/3)*(8*10^-4)
= **5.33 x 10^-4 J**

A
parallel plate capacitor has capacitance C. Determine the new
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(20) An isolated parallel-plate capacitor (not connected
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separation between the plates initially is d=1.2 mm, and
for this separation the capacitance is 3.1x10-11 F.
Calculate the work that must be done to pull the plates apart until
their separation becomes 4.5 mm, if the charge on the
plates remains constant. The capacitor plates are in a vacuum.
Show your work!!!
Also plz include a picture/description of the problem too
plz.

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