Question

The parallel plates in a capacitor, with a plate area of 7.80 cm2 and an air-filled separation of 2.70 mm, are charged by a 5.80 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 9.70 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:

(a) Potential difference between the plates is 20.84 V

(b) Initial stored energy is 4.31 x 10⁻¹¹ J

(c) Final stored energy is 15.42 x 10⁻¹¹ J

(d) Work done required is  11.11 x 10⁻¹¹ J

Explanation:

Given:

Area of parallel plate capacitor, A = 7.80 cm² = 7.80 x 10⁻⁴ m²

Initial separation between the plates, d₁ = 2.70 mm = 2.70 x 10⁻³ m

Initial potential difference between the plates, V₁ = 5.80 V

Initial Capacitance of the parallel plate capacitor is give by the relation:

C₁ = ∈₀A/d₁

Substitute the suitable values in the above equation.

$C_{1}=\frac{8.85\times10^{-12}\times7.80\times10^{-4}}{2.70\times10^{-3}}$

C₁ = 2.56 x 10⁻¹² F

Initial charge on the capacitor is given by the relation:

q₁ = C₁ V₁

Substitute the suitable values in the above equation.

$q_{1}=2.56\times10^{-12}\times5.80$

q₁ = 1.48 x 10⁻¹¹ C

Since, the capacitor is disconnected from the battery. So, the final charge of capacitor is equal to its initial charge. That is,

q₂ = q₁ =  1.48 x 10⁻¹¹ C

After disconnecting, the separation between the plates increases and its value is:

d₂ = 9.70 mm = 9.70 x 10⁻³ m

Final Capacitance of the parallel plate capacitor is give by the relation:

C₂ = ∈₀A/d₂

Substitute the suitable values in the above equation.

$C_{2}=\frac{8.85\times10^{-12}\times7.80\times10^{-4}}{9.70\times10^{-3}}$

C₂ = 0.71 x 10⁻¹² F

(a) The new potential difference is given by the relation:

V₂ = q₁/C₂

Substitute the suitable values in the above equation.

$V_{2} =\frac{1.48\times10^{-11} }{0.71\times10^{-12}}$

V₂ = 20.84 V

(b) Initial stored energy is given by the relation:

$U_{1} =\frac{1}{2} C_{1} V_{1} ^{2}$

Substitute the suitable values in the above equation.

$U_{1} =\frac{1}{2}\times2.56\times10^{-12}\times(5.8) ^{2}$

U₁ = 4.31 x 10⁻¹¹ J

(c)  Final stored energy is given by the relation:

$U_{2} =\frac{1}{2} C_{2} V_{2} ^{2}$

Substitute the suitable values in the above equation.

$U_{2} =\frac{1}{2}\times0.71\times10^{-12}\times(20.84) ^{2}$

U₂ = 15.42 x 10⁻¹¹ J

(d) Work required to separate the plates is given by the relation:

W = U₂ - U₁

W = 15.42 x 10⁻¹¹ - 4.31 x 10⁻¹¹

W = 11.11 x 10⁻¹¹ J