Answer:
0.7 mJ
Explanation:
Identify the unknown:
The work required to turn the dial from 180° to 0°
List the Knowns:
Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F
Voltage of the battery: V = 130 V
Set Up the Problem:
Energy stored in a capacitor:
U_c=1/2*V^2*C
=1/2*Q^2/C
When the dial is set at 180°:
U_c=1/2*(130)^2*350*10^-12=10^-4
Q=√2*U_c*C=4*10^-7
When the dial is set at 0°:
U_c=1/2*(4*10^-7)^2/100*10^-12
=8*10^-4 J
Solve the Problem:
ΔU_c=7*10^-4 J
=0.7 mJ
note:
there maybe error in calculation but method is correct
455pF
Given;
Turning dial from 0° to 180° changes;
capacitance from 100 to 350pF
Also, we know that the capacitance (C) of a capacitor is related to the charge(Q) on and voltage (V) across the capacitor as follows;]
Q = CV --------------------------(i)
Now, since at 180° the capacitor is connected to 130V, this means that the capacitance of the capacitor at that point is 350pF.
i.e
at V = 130V, C = 350pF = 3.50 x 10⁻¹²F
Substitute these values into equation (i) as follows;
Q = 3.50 x 10⁻¹² x 130
Q = 455 x 10⁻¹²C
Q = 455pF
Therefore, the charge of the capacitor at 180°, 350pF is 455pF.
But, since the charge on a capacitor remains unless it is discharged, the charge of the capacitor even at 0° at a disconnected battery source remains 455pF.
