The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 8.20 cm from the origin O. For each of the situations in the drawing, determine the magnitude of the net electric field at the origin.

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 8.20 cm from the origin O . For each of the situations in the drawing, determine the magnitude of the net electric field at the origin.

Given, all charges in the figure a and b are in the units of  μ  C .

The answer to the question is

7.796×10⁶ N / C  at 45.0 ° above the x axis

Explanation:

q₁ = 2.0 μ C

q₂ = -3.0 μ C

q₃ = -5.0 μ C

The principle of superposition states that the net electric field is equal to the sum of the individual electric fields

∑

Therefore   ∑ = + +

 = k × = 8.99×10⁹ N m²/ C₂× (2.0×10⁻⁶)/(8.2×10⁻² m)² = 2.674×10⁶ N / C + x

=  k × = 8.99×10⁹ N m²/ C₂× (5.0×10⁻⁶)/(8.2×10⁻² m)² = 6.685×10⁶ N / C + y

=k × = 8.99×10⁹ N m²/ C₂× (3.0×10⁻⁶)/(8.2×10⁻² m)² = 4.011×10⁶ N / C + x

2.674×10⁶ N / C + x + 6.685×10⁶ N / C + y + 4.011×10⁶ N / C + x =

= 6.685×10⁶ N / C + x+ 6.685×10⁶ N / C + y

Hence = √(( 6.685×10⁶)²+(6.685×10⁶)²) = 7.796×10⁶ N / C

θ = tan ⁻¹ (6685008.923/6685000)  = 45.0 °

Therefore 7.796×10⁶ N / C  at 45.0 ° above the x axis

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