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

The surface charge density on the plate,
`\sigma=7.5* 10^(-5)\ C/m^2`

Step-by-step explanation:

It is given that,

Area of parallel plate capacitor,
`A=2* 10^(-3)\ m^2`

Separation between the plates,
`d=5* 10^(-2)\ mm`

Electric field between the plates,
`E=8.5* 10^(6)\ V/m`

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

`E=(\sigma)/(\epsilon)`

Where

`\sigma` = surface charge density

`\sigma=E* \epsilon`

`\sigma=8.5* 10^(6)\ V/m* 8.85* 10^(-12)\ C^2/Nm^2`

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

`\sigma=7.5* 10^(-5)\ C/m^2`

Hence, this is the required solution.