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.

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

correct it is c.) b b