Charging a capacitor at a constant rate

Precautions:

The resistance R must be large so that the current, i.e the rate of charging is small. Otherwise charging is completed too rapidly to be observed.

Theories:

From definition, the capacitance C of a capacitor is found from

C = Q/V

where Q is the charge stored on the capacitor and V is the potential difference across it.

 Q=CV

If a capacitor is charged up at a constant rate, i.e., where I is a constant,

Then  is also constant.

Hence the potential difference across the capacitor increases linearly with time.

Measurements:

Current, which is kept constant = 80µA

From the graph, a straight line passing through origin is obtained. It shows that the p.d across the capacitor is directly proportional to the time. When the time increases, the p.d across the capacitor increases linearly.

Discussion:

• When the capacitor is charging up, the current decreases with time. By reducing the resistance of the variable resistor R, the charging current can be kept constant. As the current is constant, the charge stored in the capacitor directly proportional to the time, i.e., Q =IT. Since the p.d across the capacitor is also directly proportional to the time, the charge on the capacitor is directly proportional to the p.d across it. i.e. QV.
• From the graph, the slope of the straight line = (5-1)V/(40-8)s= 0.125Vs-1

 since

    0.125 = 80µA/C

            C=6.4 X 10-4CV-1

• Sources of errors:

- There is leakage of charge from the capacitor.

- Error in keeping the constant current.

- There is internal resistance of the cell.

- Non-zero value of variable resistance when it is set "minimum".

Conclusion:

By charging the capacitor at a constant rate, the charge on a capacitor is directly proportional to p.d. applied across it