a parallel plate capacitor, 10µF with air between its plate is connected to a 50 volt source and then disconnected.

a)What is the charge on the capacitor and the potential difference across it.

the potential difference across it is 50 v. Odd question.

Charge: Capacitance= charge/potential

charge= capacitance*voltagepotential

To find the charge on the capacitor, we can use the formula:

Q = C * V

Where:
Q is the charge on the capacitor,
C is the capacitance of the capacitor,
V is the potential difference across the capacitor.

In this case, the capacitance (C) is given as 10µF (microfarads), which we need to convert to farads. Since 1µF = 10^-6 F, we have:

C = 10µF = 10 * 10^-6 F = 10^-5 F

The potential difference (V) is given as 50 volts.

Substituting the values into the formula, we have:

Q = (10^-5 F) * (50 V)
Q = 5 * 10^-4 Coulombs

Therefore, the charge on the capacitor is 5 * 10^-4 Coulombs.

Now, after the capacitor is disconnected from the source, the charge on it remains constant. However, the potential difference across the capacitor becomes zero, as no voltage is applied.

So, the potential difference across the capacitor is 0 volts.