A parallel plate capacitor of capacitance C_0 has plates of area A with separati
ID: 1528544 • Letter: A
Question
A parallel plate capacitor of capacitance C_0 has plates of area A with separation d between them. When it is connected to a battery of voltage V_0, it has charge of magnitude Q_0 on its plates. The plates are pulled apart to a separation 2d while the capacitor remains connected to the battery. After the plates are 2d apart, the capacitance of the capacitor and the magnitude of the charge on the plates are 1/2 C_0, 1/2 Q_0 1/2 C_0, Q_0 C_0, Q_0 2C_0, Q_0 2C_0, 2Q_0 Determine the equivalent capacitance of the combination shown when C = 12 pF.Explanation / Answer
According to the given data:
capacitance of a parallel plate capacitor is C = (epcilon 0)*A/d
here A - area
d- seperation
A is constant
d becomes 2d
so,
C1/C2 = d2/d1
C1/c2 = 2d/d =2
C2 = (1/2)C1
C(new) = (1/2)C0
and charge Q= C*V
Q1/Q2 = C1/C2 (because V is same)
Q1/Q2 = 2
Q2 = (1/2) Q1
Q(new) = (1/2) Q0
2. C= 12 pF
here 2 capacitors which are middle in parallel
so C' = C+C = 24 pF
Now total capacitors in series
1/C" = 1/C + 1/C' + 1/C
1/C" = 1/12 + 1/24 + 1/12
C" = 4.807 pF