Question

The parallel plates in a capacitor, with a plate area of 7.10 cm2 and an air-filled separation of 2.20 mm, are charged by a 4.80 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 6.50 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)14.17V

b)32.8 x $10^-^1^2$J

c)96.9x $10^-^1^2$J

d) -64x $10^-^1^2$J

Explanation:

Given:

Area 'A'=7.10cm² =>7.1 x $10^-^4$m²

voltage '$V_o$'=4.8 volt

$d_o$ = 2.20mm => 2.2 x $10^-^3$m

$d_1$ = 6.50mm => 6.5 x $10^-^3$m

a) Capacitance $C_o$ before push is given by:

$C_o$ = εA/$d_o$ =>$\frac{(8.85*10^-^1^2)(7.1*10^-^4)}{2.2*10^-^3}$

$C_o$ =   2.85 x $10^-^1^2$ F

$q_o$=$C_o$$V_o$=> 2.85 x $10^-^1^2$ x 4.8

$q_o$=1.37 x $10^-^1^1$ C

Capacitance $C_1$ after push is given by:

$C_1$ = εA/$d_1$ =>$\frac{(8.85*10^-^1^2)(7.1*10^-^4)}{6.5*10^-^3}$

$C_1$ =   9.66 x $10^-^1^3$F

$q_o$=$q_1$

$q_1$=$C_1$$V_1$

Therefore, the potential difference between the plates

$V_1$ = 1.37 x $10^-^1^1$ / 9.66 x $10^-^1^3$ =>14.17V

b) $U_i=\frac{1}{2}C_oV_o^2 => \frac{1}{2} (2.85*10^-^1^2)(4.8^2)$

$U_i=$ 32.8 x $10^-^1^2$J

c)$U_f=\frac{1}{2}C_1V_1^2 => \frac{1}{2} (9.66*10^-^1^3)(14.17^2)$

$U_f$= 96.9x $10^-^1^2$J

d) the work required to separate the plates is given by:

workdone=  $U_i$-$U_f$=> 32.8 x $10^-^1^2$J- 96.9x $10^-^1^2$J

W≈ -64x $10^-^1^2$J