A(n) 12 µF air-filled capacitor is charged to a potential difference of 3473 V.

Question

A(n) 12 µF air-filled capacitor is charged to a
potential difference of 3473 V.

A) What is the energy stored in it?
Answer in units of J

B) If the voltage source remains connected to the
capacitor, but the air is replaced by a layer of
plastic with dielectric constant of 4.9, what is
the new value of energy stored in it?
Answer in units of J

C) A 35 pF capacitor is charged to 2 kV and
then removed from the battery and connected
in parallel to an uncharged 50 pF capacitor.
What is the new charge on the second ca-
pacitor?
Answer in units of nC

Explanation / Answer

a) E=1/2cV^2=72.37 J

b) New value of C=12x10^(-6)x4.9

so E=354.61J

c) just after removing potential difference across 35pF capacitor= 2kV

let the charges be q1 and q2 on 35pF and 50pF respectively then

q1/35=q2/50

and q1+q2=(35x10^(-12)x2000) (charge conservation)

solving we get

q1=2.88x10^(-8)

q2=4.12x10^(-8)

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