Question

The parallel plates in a capacitor, with a plate area of 5.40 cm2 and an air-filled separation of 4.30 mm, are charged by a 4.60 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.10 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.

Answer 1

Answer:

Explanation:

given,

plate area = 5.40 cm²

separation = 4.30 mm

Capacitance = $\dfrac{\epsilon A}{d}$

                     = $\dfrac{8.85\times 10^{-12}\times 5.4\times 10^{-4}}{4.3 \times 10^{-3}}$

                    =$1.11 \times 10^{-12} F$

Potential difference applied = 4.6 V

charge stored in the capacitor = $4.6 \times 1.11 \times 10^{-12}$

                                                   =$5.106 \times 10^{-12}$

Now when plates are moved new capacitance

Capacitance = $\dfrac{\epsilon A}{d}$

                     = $\dfrac{8.85\times 10^{-12}\times 5.4\times 10^{-4}}{8.10 \times 10^{-3}}$

                    =$0.59 \times 10^{-12} F$

voltage =$\dfrac{5.106 \times 10^{-12}}{0.59 \times 10^{-12}}$

potential applied = 8.65 V

Initial energy stored =$\dfrac{1}{2}C_1V^2$

                                  =$\dfrac{1}{2}\times 1.11 \times 10^{-12} 4.6^2$

                                  = $1.174\times 10^{-11} J$

Final energy stored  =$\dfrac{1}{2}C_1V^2$

                                  =$\dfrac{1}{2}\times 0.59 \times 10^{-12} 8.65^2$

                                  =$2.21\times 10^{-11} J$

Work done = final energy - initial energy

                 = $2.21\times 10^{-11} - 1.174\times 10^{-11}$

                 =  $1.036\times 10^{-11}$