Respuesta :
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 .
Answer:
Explanation:
Area of plates, A = 9.50 cm²
separation, d = 3.40 mm
Voltage, V = 7.60 V
new separation, d' = 9.5 mm
The formula of the parallel plate capacitance is given by
[tex]C = \frac{\epsilon_{0}A}{d}[/tex]
[tex]C = \frac{8.854\times 10^{-12}\times 9.5\times 10^{-4}}{3.4\times 10^{-3}}[/tex]
C = 2.44 x 10^-12 F
initial charge, q = C x V
q = 2.44 x 10^-12 x 7.6 = 1.88 x 10^-11 C
(a)
As the battery is disconnected so the charge remains same..
The new capacitance, C' is given by
[tex]C' = \frac{\epsilon_{0}A}{d'}[/tex]
[tex]C' = \frac{8.854\times 10^{-12}\times 9.5\times 10^{-4}}{9.5\times 10^{-3}}[/tex]
C' = 8.854 x 10^-13 F
Let the new potential difference is V'
V' = q / C'
V' = (1.88 x 10^-11) / (8.854 x 10^-13) = 21.23 V
(b) Initial energy
U = 0.5 x CV²
U = 0.5 x 2.44 x 10^-12 x 7.6 x 7.6 = 7.05 x 10^-11 J
(c) Final energy
U' = 0.5 x C' x V'²
U' = 0.5 x 8.854 x 10^-13 x 21.23 x 21.23 = 1.995 x 10^-10 J
(d) Work done = change in potential energy
W = U' - U
W = 1.995 x 10^-10 - 7.05 x 10^-11
W = 12.9 x 10^-11 Joule
