Question

Each plate of a parallel-plate air-filled capacitor has an area of 2×10−3 m2, and the separation of the plates is 5×10−2 mm. An electric field of 8.5 ×106 V/m is present between the plates. What is the surface charge density on the plates? (ε 0 = 8.85 × 10-12 C2/N · m2)

Answer 1

Answer:

The surface charge density on the plate, $\sigma=7.5\times 10^{-5}\ C/m^2$

Explanation:

It is given that,

Area of parallel plate capacitor, $A=2\times 10^{-3}\ m^2$

Separation between the plates, $d=5\times 10^{-2}\ mm$

Electric field between the plates, $E=8.5\times 10^{6}\ V/m$

We need to find the surface charge density on the plates. The formula for electric field is given by :

$E=\dfrac{\sigma}{\epsilon}$

Where

$\sigma$ = surface charge density

$\sigma=E\times \epsilon$

$\sigma=8.5\times 10^{6}\ V/m\times 8.85\times 10^{-12}\ C^2/Nm^2$

$\sigma=0.000075\ C/m^2$

$\sigma=7.5\times 10^{-5}\ C/m^2$

Hence, this is the required solution.