Question

ABC is an equilateral triangle of side 1m. Charges of -1 μC and 1 μC are placed respectively at points B and C. The electric field at point A is _______. Single choice.

(1 Point)

9x10^2 MKS

18 x 10^2

9x10^3 MKS parallel to BC

9x10^3 MKS prallel to CB


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Answer 1

The value of the electric field is 9x10^3 MKS and is parallel to the direction of BC.

We are given that:

• The length of the side = 1m
• Charges = -1 μC and 1 μC
• To Find: The electric field at point A

Solution:

• Electric field E1 = k x 1-^-6 / 1 = 10^-6 K
• Electric field E2 = k x 10^-6 / / 1 = 10^-6 K
• E(net) = 2[10^-6cos(60°)] = 10^-6 K
• E(net) = 10^-6 x 9 x 10^9  = 9 x 10^3
• The direction of electric field is parallel to BC.

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