A given capacitor is connected to a 12.0 V battery, and when fully charged store

Question

A given capacitor is connected to a 12.0 V battery, and when fully charged stores 3.67 x 10-4 J of energy. If the plate separation of the capacitor is decreased by a factor of 2, with the capacitor remaining connected to the battery, how much energy is stored in the capacitor in the new configuration?

S(k*q/r) = 0 as electric potential is scalar quantity.
so,
k* S(q/r) = 0
as k is constant
so,
S(q/r) = 0
let third charge be q
let side of triangle be a
so, distance of 2.00 nC and - 5.00 nC from thier midpoint = a/2
also, distance of third vertex from that point = v3*a/2
inputing values:-
q/(v3*a/2) + (2 e-9)/(a/2) + (-5 e-9)/(a/2) = 0
multiplying throughout by (a/2) (to get rid of it in denominator) and solving for q:-
q/v3 = 3 e-9
so,
q = v3 * 3 e-9
q = 5.1961 x 10-9 C ˜ 5.20 nC

In this answer where does the 5.1961x10-9C come from to get the answer of 5.20nC

Explanation / Answer

The capacitance is given by: So if the plate separation is halved the capacitance is doubled. And the energy is given by: So when the capacitance is doubled the energy is doubled. 2*3.67*104 =7.34*104 J

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