# Determine the charge on the plates before and after immersion.

A parallel-plate capacitor in air has a plate separation of 1.63 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.

(a) Determine the charge on the plates before and after immersion.
 before pC after pC

(b) Determine the capacitance and potential difference after immersion.
 Cf = F ΔVf = V

(c) Determine the change in energy of the capacitor.
nJ

asked Jan 27, 2012 in Physics

The charge will be the same before and after the immersion beacuse the liquid is an insulator. You can calculate this charge by using the formula for the parallel plate capacitor:

C = E*A / d  (where C = capacitance, E = permitivity of free space, A = area, d = distance between plates)

Now replace C with the formula for capacitance, which is:

C = Q / V  (where C = capacitance, Q = charge, and V = voltage)

rewrite the first equation now:

Q / V = E*A / d

and then solve for the charge Q:

Q = (E*A * V) / d

Now just plug in the values and solve for Q:

E = 8.85*10^-12 C^2 / N*m^2

A = 25 * 10^-4 m

d = 1.63 * 10^-2 m

V = 255 V

Q = ((8.85*10^-12 C^2 / N*m^2) * (25 * 10^-4 m) *( 255 V)) / (1.63 * 10^-2 m)

Q = 3.46*10^-10 C  =    346 pC    (for both parts of A, because it is immersed in an insulator)

For Part B:

You can calculate the capacitance byusing the formula for the parallel plate capacitor and multiplying by the kappa for water:

Cf = k*E*A / d  (where Cf = final capacitance, k = kappa for water,  E = permitivity of free space, A = area, d = distance between plates)

k = 80

E = 8.85*10^-12 C^2 / N*m^2

A = 25 * 10^-4 m

d = 1.63 * 10^-2 m

plug in these values and solve for C.

Cf = 80 * (8.85*10^-12 C^2 / N*m^2 ) * (25 * 10^-4 m) / (1.63 * 10^-2 m)

Cf = 1.09*10^-10 F

similarly, Vf can be found by using the ralation that C = Q/V

and so

Vf = (Q*d) / (k*E*A)

at this point the Vf can be solved however, you could replace Q with (E*A*V / d) based on the equation C = Q/V, and so the above equation would simplify to:

Vf = Vi / k

Vf = 255 V / 80  =   3.19 V

Part C, the change in energy of the capacitor is

change in energy of capacitor =  Uf - Ui

with Uf = (1/2)*Cf*Vf^2

and Ui = (1/2)*Ci*Vi^2

where Ci = initial conductance, which is Q/ Vi

Ci = Q/ Vi =  3.46*10^-10 C / 255V  = 1.35*10^-12 F

Vi = 255V

Cf = 1.09*10^-10 F  (from part B)

Vf = 3.19 V   (from part B)

plugging in the equation:

Uf - Ui =  (1/2)*Cf*Vf^2  -   (1/2)*Ci*Vi^2

= (1/2)*(1.09*10^-10 F)*(3.19 V)^2  -   (1/2)*(1.35*10^-12 F)*(255V)^2

= -4.33*10^-8 J   =   -43.3 nJ

change in energy of capacitor is =  -43.3 nJ
answered Jan 28, 2012 by ~Expert~ (3,856 points)
selected Jan 28, 2012 by apush