Question

if the potential of a capacitor having capacity 8pF is increased from 10v to 20 v then increase in the energy will be​

Answer 1

Answer:

Explanation:

30 volts into 10n

Answer 2

Given,

The capacitance of the capacitor = c = 8pF = 8 × 10⁻¹² F

Initial potential = $V_{i}$ = 10 V

Final potential = $V_{f}$ = 20 V

To find,

The change (increase) in energy with an increase in the potential.

Solution,

We know that,

∴ The energy of a capacitor = U =  $\frac{1}{2} cV^{2}$

                                                        where, c= capacitance and V = potential

Thus,

Change in energy (ΔU) = $U_{f}- U_{i}$  

                                       = $\frac{1}{2} c V_{f}^{2} - \frac{1}{2} c V_{i}^{2}$

                                       = $\frac{1}{2}c (V_{f}^{2} - V_{i}^2)$

                                       = $\frac{1}{2} * 8 * 10^{-12}* (20^2-10^2)$

                                       = $4 * 10^{-12} *(400-100)$

                                       = $4*10^{-12}*300$

                                       = $12* 10^{-10}$ J

                                       =  $1.2 * 10^{-9}$ J

Hence,  The change (increase) in energy is  $1.2*10^{-9}$ Joules.

(Remember that the energy will be in Joules(J) if and only if when the capacitance is measured in Farad(F), and the voltage is measured in Volt(V). In this question, the capacitance is given in pF which means we have to convert 'pF' to 'F' to get the energy in Joules.)

∴  1 pF = 10⁻¹² F