Question

A set of charged plates have anarea of 5.10*10^-3 m^2 andseparation 1.42*10^-5 m. Howmuch charge must be placed onthe plates to create a potentialdifference of 125 V across them?(The answer is *10^-7 C. Just fill inthe number, not the power.)

Answer 1

Given data:

Area of plates:

$A=5.10\times10^{-3}\text{ m}^2$

Separation between the plates:

$d=1.42\times10^{-5}\text{ m}$

Potential difference:

$V=125\text{ V}$

The capacitance of the capacitor is given as,

$C=\frac{A\epsilon_{\circ}}{d}$

Here, ε_o is the permittivity of the free space.

Substituting all known values,

$\begin{gathered} C=\frac{5.10\times10^{-3}\times8.85\times10^{-12}}{1.42\times10^{-5}} \\ =3.178\times10^{-9}\text{ F} \end{gathered}$

The charge on the capacitor is given as,

$Q=CV$

Substituting all known values,

$\begin{gathered} Q=3.178\times10^{-9}\times125 \\ =3.9725\times10^{-7}\text{ C} \end{gathered}$

Therefore, the charge on the plates is 3.9725×10^-7 C.