Question

...Continuation of question involving material and dielectric constant. SEE BELOW...

B) A 100pF capacitor is charged to a potential difference of 50V, and the charging battery is disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to 35 V, what is the capacitance of this second capacitor? (6)

C) Two capacitors, of 2.0 and 4.0μF capacitance are connected in parallel across a 300V potential difference. Calculate the total energy stored in the capacitors? (2)

(see last ques above)

Answer 1

for b)
First find the charge on the first capacitor
q=CV = 100pf (50v)=5000pC
Then, considering the second condition (the capacitor is then connected in parallel with a second initially uncharged capacitor):

qtot(initial)=q1(initial)+q2(initial)
our q1(initial) will be the 5000pC

qtot(i) = 5000pC + 0 = 5000pC
qtot(i)=qtot(f)
5000pC= q1(f)+q2(f)
5000pC = C1V1+ C2V2

Since the arrangement is in parallel, the pot. difference across C1 is the same across C2. Let's denote it as Vf. Hence, V1=V2=Vf

5000pC = Vf(C1+C2)
5000pC=35(100pf+C2)
C2=42.9pf


for c)
W1 = 1/2 CV^2 = 1/2 (2x10^-6)(300)^2= 0.09 joules
W2 = 1/2 (4x10^-6)(300)^2 = 0.18 joules