A parallel plate capacitor with capacitance C has plates of area A and separation d.
If the area of the plates is increased to 2A and their separation reduced to d/2 the capacitance will be
C = εoA/d C' = εo2A/d/2 = 4εoA/d
A 5 pF capacitor is charged by connecting to a 10 V battery.
The charge stored is
Q = CV = 5 x 10 pC
A parallel plate capacitor with air gap has capacitance 4 pF. A dielectric of dielectric constant 2 is pushed halfway into the gap as shown.
The new capacitance is
Pushing in the dielectric doubles the capacitance of half of the capacitor. The capacitor is now like two capacitors in parallel.
This has a total capacitance = 6 pF
A capacitor is charged to 6 V and then isolated from the battery. The plates are then pulled apart so their separation is twice as big.
The PD between the plates will now be
V = Q/C
Q is constant and C is halved
V is therefore 2x
A capacitor is connected to a 6 V battery then disconnected and connected across and identical capacitor.
The PD across the second capacitor after they are connected will be
When connected the charge will be shared equally to the capacitors.
The charge on each will be 1/2 so the PD will be 1/2
A capacitor is charged to 6V then isolated from the battery. The plates are then pulled apart so their separation is twice as big.
If the energy stored before was E, then the energy stored after pulling apart is
The charge stays the same and the capacitance will be 1/2 so the PD will x2.
E = 1/2 QV so if V is x2, E will be x2
A 10 pF capacitor is connected to a 6V battery. The plates of the capacitor are now moved sideways so the area of overlap is 1/2.
The Charge on the capacitor will be
The PD will remain constant as the battery is still connected.
The capacitance will 1/2 so becomes 5 pF
Q = CV = 5 x 6 pC
In the diagram each capacitor has capacitance C
The total capacitance is
The two series capacitors have total capacitance = C/2
The parallel combination = C/2 + C/2 = C
1/The total = 1/C + 1/C
A parallel plate capacitor of capacitance 10 pF is charged by connecting to a 6 V battery. Plates are now pushed closer together. Which of the following statements is true
The capacitance will increase so more charge will be required to make the PD between the plates = 6V
A 2 pF capacitor is connected in series with a 4 pF capacitor.
The total capacitance is
Capacitors are in series so 1/CT = 1/C1 + 1/C2 = 1/2 + 1/4 = 3/4 CT = 4/3 pF
The voltmeter in the circuit below measures the PD across the resiator as it changes with time as shown in the input graph.
Which graph represents the PD across the resistor?
Which diagram shows the correct arrangement of diodes for a full wave rectifier