An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 7.0 V battery, with the positive and negative terminals connected in the same manner as before.

1 μC extra charge will be flow here

Explanation:

Given data

battery V1 = 4.0 V

flows Q1 = 6.0 μC

replace battery V2 = 7.0 V

to find out

what happen if we replace battery

solution

we apply here principal of capacitor

that is Q directly proportional voltage

so we say Q2/Q1 = V2 / V1

put all value here

Q2/Q1 = V2 / V1

Q2/6 = 7 / 6

Q2 = 7

so we see here 7 μC will be flow

and Q = Q2 - Q1 = 7 - 6 = 1 μC

so we also say that 1 μC extra charge will be flow here