Question

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a 23.0-V potential difference is applied to these plates, calculate the following.
(a) Calculate the electric field between the plates.


(b) Calculate the surface charge density.

(c) Calculate the capacitance.


(d) Calculate the charge on each plate.

Answer 1

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

$area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v$

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

$E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m$

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

$8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}$

c.

from the expression for the capacitance

$C=eA/d$

if we substitute values we arrive at

$C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF$

d. To calculate the charge on each plate, we use the formula below

$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$