The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With the dial set at 180°, the capacitor is connected to a 130 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned to 0°. What is the charge on the capacitor now?

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.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.