Answer:
Explanation:
capacitance of capacitor = ε₀ A / d
ε₀ = 8.85 x 10⁻¹² , A is area of plate and d is plate separation .
= 8.85 x 10⁻¹² x 9.5 x 10⁻⁴ / 3.4 x 10⁻³
C = 24.73 x 10⁻¹³
capacitance of capacitor after increase in plate separation
= 8.85 x 10⁻¹² x 9.5 x 10⁻⁴ / 9.5 x 10⁻³
= 8.85 x 10⁻¹³
initial charge = capacitance x potential
= 24.73 x 10⁻¹³ x 7.6 C
potential difference after increased separation
= initial charge / increased capacitance
= 24.73 x 10⁻¹³ x 7.6 / (8.85 x 10⁻¹³)
= 21.23 V .
b ) initial stored energy
= 1/2 C V²
= .5 x 24.73 x 10⁻¹³ x 7.6²
= 714.2 x 10⁻¹³ J
c ) final stored energy
1/2 C V²
= .5 x 8.85 x 10⁻¹³ x 21.23 ²
= 1994.4 x 10⁻¹³ J
d ) work done in separation of plate
=2 x increase in stored energy
= 2 x (1994.4 - 714.2 ) x 10⁻¹³ J
= 2560.4 x 10⁻¹³ J .
