A 500f parallel plate capacitor is connected to a 120 v...

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconnected without loss of any charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) what would the voltmeter read if (i) the plate separation was doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

Transcript

00:01 Hi, in this question we are given with a capacitance of the capacitor is 5 microfarad and the voltage or the potential difference between the capacitor plate is 12 volt.

00:09 So, the charge stored in the capacitor c is given by q is equal to c into v that is 12 into 5 microfarad which is equal to 60 micro coulombs.

00:23 Then we have the voltage reading is 12 units itself.

00:27 So, in the first question the voltage is connected across the capacitor.

00:31 So, it is assumed that so voltage is connected across the capacitor and it is assumed that no charge.

00:42 So, there is no loss of charge no loss of charge and so in this is in case of ideal voltmeter but in practical voltmeter this not the case.

00:59 So, the reading of the voltmeter v is equal to 12 volt as there is no loss of charge no charge loss.

01:12 So, we have the formula q is equal to cv.

01:15 So, q will be equal to 60 micro coulomb.

01:23 So, therefore, the potential will be equal to 12 volts.

01:29 In the second case we have to find the voltmeter reading.

01:34 So, voltmeter reading if the plate suppression.

01:38 So, if the plate suppression d.

01:42 So, d is doubled.

01:46 So, plate suppression is doubled.

01:48 Now we have the capacitance is given by.

01:52 So, we have c is equal to epsilon naught a divided by d where epsilon naught is the absolute permittivity of the prefix space.

02:01 So, as the suppression is doubled.

02:04 So, this is doubled which implies for another capacitance epsilon naught a.

02:10 So, these are constant only suppression is doubled.

02:13 So, this becomes 2d.

02:15 Therefore, this c dash will be equal to it is c by 2 that is the capacitance is.

02:21 So, capacitance is halved.

02:26 So, it is half of the original value.

02:30 Therefore, we have c is equal to.

02:34 So, c will be equal to 5 micro farad then c dash will be equal to 2 .5 micro farad.

02:43 Then in the second case.

02:45 So, we have also here.

02:50 So, in this case q remains constant.

02:53 So, q remains constant, but we have q is equal to c into v...

Question

Please give Ace some feedback