Question

A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? 2Q 4Q Q Q/2 Request Answer Part B - Potential difference in a parallel-plate capacitor

Answer 1

2Q

Step-by-step explanation:

The expression for the electric charge stored in a capacitor is given as,

Q = CV..................... Equation 1

Where Q = Electric charge, C = Capacitance of the capacitor, V = Potential difference applied across the plates of the capacitor.

make C the subject of the equation

C = Q/V............... Equation 2

Given: Q = Q, V = 9 V

Substitute into equation 2

C = Q/9 F.

If it is connected to an 18 V battery instead,

Q = CV

Given: C = Q/9 F, V = 18 V

Q = (Q/9)(18)

Q = 2Q.

Hence the charge on the capacitor = 2Q.

Answer 2

Part A:

`2Q`

Part B:

`V_a_b=(Q)/(C)`

Step-by-step explanation:

The capacitance is given by:

`C=(Q)/(V_a_b)`

Where:

`Q=Stored\hspace{3}electric\hspace{3}charge\\V_a_b=Potential\hspace{3}difference\\C=Capacitance`

So:

`Q=C V_a_b`

In the first case:

`V1_a_b=9`

And in the second case:

`V2_a_b=18`

or:

`V2_a_b=2*V1_a_b=2*9=18`

C is a constant so:

`Q_1=C*V1_a_b\\\\so\\V1_a_b=(Q_1)/(C) \\ \\\\and\\\\Q_2=C*V2_a_b`

Therefore:

`Q_2=C*2V1_a_b\\\\Q_2=C*(2Q_1)/(C) \\\\Q_2=2*Q_1`

Part B:

To find thepotential difference in a parallel-plate capacitor just isolate
`V_a_b` from the capacitance equation:

`V_a_b=(Q)/(C)`