# Two Charged Capacitors in Parallel

1. Feb 16, 2010

### littlebilly91

1. The problem statement, all variables and given/known data

Two large area parallel plate capacitors, labeled P and N, are connected as shown in the figure below. The charge on each plate is indicated in the figure, in μC.

I. The capacity of N (on the right) is 27.5 μF. Calculate the capacity of P.
1.34×10-5 F (Correct)

The plate separation of capacitor of N (on the right) is doubled. Calculate the new value of the charge on P.

2. Relevant equations

q=CV
total initial charge=total final charge
C=$$\epsilon$$A/d

3. The attempt at a solution

$$C_{Ni}$$=$$C_{Nf}$$/2
because the distance has doubled (C=$$\epsilon$$A/d)

$$q_{Ni}$$+$$q_{Pi}$$ = $$q_{Nf}$$+$$q_{Pf}$$

$$q_{Ni}$$+$$q_{Pi}$$ = $$C_{Nf}$$V+$$q_{Pf}$$

$$q_{Ni}$$+$$q_{Pi}$$ = $$q_{Pf}$$*$$C_{Nf}$$/($$2*C_{Pf}$$)+$$q_{Pf}$$
using q=CV because the voltage drop will be the same between the two final capacitors

then I solved for qPf

File size:
969 bytes
Views:
78