Question

A square conducting plate 51.0 cm on a side and with no net charge is placed in a region, where there is a uniform electric field of 81.0 kN/C directed to the right and perpendicular to the plate. (a) Find the charge density (in nC/m2) on the surface of the right face of the plate. -716850 nC/m2 (b) Find the charge density (in nC/m2) on the surface of the left face of the plate. 716850 nC/m2 (c) Find the magnitude (in nC) of the charge on either face of the plate. nC

Answer 1

a)
`\sigma = 716.85nC/m^3`

b)
`\sigma = -716.85nC/m^3`

c) i)
`Q = 198.157 nC`

ii)
`Q = 198.157 nC`

Step-by-step explanation:

To find the charge density on each face, let's use the formula: E=σ/εo

Where, E, electric field = 81.0 kN/C

εo =
`8.85*10^-^1^2`

Thus, solve for σ

`\sigma = 81.0*10^3 * 8.85*10^-^1^2`

`\sigma = 7.1685*10^-^7 C/m^3` or
`\sigma = 716.85nC/m^3`

In charge density, the left face is negative while on the right face it will be positive.

Therefore,

Charge density on the each face =

`- 716.85nC/m^3 and 716.85nC/m^3`

C) We'll first find the area of the square plate.

`A = (51.0*10^-^2)^2 = 0.2601m^2`

Use the formula below to find the magnitude of the charge on each surface of the plate:

On the right surface:

`Q = A\sigma`

`Q = 0.2601 * 761. 85`

`Q = 198.157 nC`

On the left surface:

`Q = A\sigma`

`Q = 0.2601 * -761. 85`

`Q = -198.157 nC`