Question

A parallel-plate capacitor has an area A = 3.00 cm2
and a plate separation d =1 mm. (a) Find
its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a
6.00-V battery? Calculate (c) the charge density on the positive plate, assuming the density is
uniform, and (d) the magnitude of the electric field between the plates

Answer 1

As per the given question,

Plate area of the capacitor $(A)=3.0 cm^2=3.0\times 10^{-4}m^2$

Distance between the plates$(d)= 1mm=1\times 10^{-3}m$

Permittivity of free space $(\epsilon_o) =8.85\times 10^{-12}F/m$

i)

We know that the capacitance of the capacitor $=\frac{\epsilon_o A}{d}$

Now, substituting the values,

$C=\frac{8.85\times 10^{-12}\times 3\times 10^{-4}}{1\times 10^{-3}}F$

$\Rightarrow C=26.46\times 10^{-13}F$

$\Rightarrow C= 2.646\times 10^{-12}F$

ii) V=6.0 Volt

Hence charge stored in the capacitor $(q)= cV =2.646\times 10^{-12}\times6 =15.876\times 10^{-12}C$

iii) Charge density on the positive plate of the capacitor $=q/A$

Substituting the values,

$=\frac{15.876\times 10^{-12}}{3.0\times 10^{-4}}C/m^2$

$=5.292\times 10^{-8}C/m^2$

iv) Magnitude of the electric field between the plates $E=\frac{-\Delta V}{d}$

$=\frac{6}{3\times 10^{-4}}N/C =2\times 10^{4}N/C$