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

Answer:

E = 5.65 N/C

Explanation:

Given data:

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

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

charge density $\sigma_1 creates the electric field E_x = \frac{\sigma_1}{x}$

And A charge density $\sigma_2$ creates the electric field $E_y = \frac{\sigma_1}{y}$

The electric field at point (x,y) is

$E = E_y + E_x$

  $=\frac{\sigma_2}{y} e_y + E_x = \frac{\sigma_1}{x}e_x$

where $e_y and e_x$ are vectors

After solving we get

$E = \sqrt{(\frac{\sigma_1}{y})^2+(\frac{\sigma_1}{y})^2}$

$E = \sqrt{(\frac{60\TIMES 10^{-12}}{2\times 8.85\times 10^{-12}})^2+(\frac{80\times 10^{-12}}{2\times 8.85\times 10^{-12}})^2}$

E = 5.65 N/C

Answer 2

Answer:

5.648 N/C

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 = $\frac{60\times10^{-12}}{2\times8.85\times10^{-12}}$

or

E₁ = 3.389 N/C

and, due to charge q₂

E₂ = $\frac{-80\times10^{-12}}{2\times8.85\times10^{-12}}$

or

E₂ = 4.519 N/C

Now,

The resultant electric field = $\sqrt{E_1^2+E_2^2}$

or

The resultant electric field = $\sqrt{3.389^2+4.519^2}$

or

The resultant electric field = $\sqrt{11.485321+20.421361}$

or

The resultant electric field = 5.648 N/C