Question

Two infinite, uniformly charged, flat surfaces are mutually perpendicular. One of the sheets has a charge density of +60 pC/m2, and the other carries a charge density of −80 pC/m2. What is the magnitude of the electric field at any point not on either surface?

Answer 1

E = 5.65 N/C

Step-by-step explanation:

Given data:

Charge density
`\sigma_1 = +60 pC/m^2`

`\sigma_2 = -80pC/m^2`

charge density
`\sigma_1 creates the electric field E_x = (\sigma_1)/(x)`

And A charge density
`\sigma_2` creates the electric field
`E_y = (\sigma_1)/(y)`

The electric field at point (x,y) is

`E = E_y + E_x`

`=(\sigma_2)/(y) e_y + E_x = (\sigma_1)/(x)e_x`

where
`e_y and e_x` are vectors

After solving we get

`E = \sqrt{((\sigma_1)/(y))^2+((\sigma_1)/(y))^2}`

`E = \sqrt{((60\TIMES 10^(-12))/(2* 8.85* 10^(-12)))^2+((80* 10^(-12))/(2* 8.85* 10^(-12)))^2}`

E = 5.65 N/C

Answer 2

5.648 N/C

Step-by-step explanation:

Given:

q₁ = 60 pC/m² = 60 × 10⁻¹² C/m²

q₂ = -80 pC/m² = - 80 × 10⁻¹² C/m²

Now,

Electric field is given as:

E =
`\frac{\textup{q}}{2\epsilon_0}`

ε₀ = Permittivity of Free Space

thus, due to charge q₁

E₁ =
`(60*10^(-12))/(2*8.85*10^(-12))`

or

E₁ = 3.389 N/C

and, due to charge q₂

E₂ =
`(-80*10^(-12))/(2*8.85*10^(-12))`

or

E₂ = 4.519 N/C

Now,

The resultant electric field =
`√(E_1^2+E_2^2)`

or

The resultant electric field =
`√(3.389^2+4.519^2)`

or

The resultant electric field =
`√(11.485321+20.421361)`

or

The resultant electric field = 5.648 N/C