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olya-2409 [2.1K]
3 years ago
9

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?
Physics
2 answers:
Charra [1.4K]3 years ago
6 0

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

andreyandreev [35.5K]3 years ago
5 0

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

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