Problem: A. What is the magnitude of the electric force on charge A in the figure?F=______NB. What is the direction of the electric force on charge A in the figure? Choose the best answer.(a) to the left(b) to the right(c) the force is zero

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FREE Expert Solution

A. Electric force at point A is due to the charges at point B and C. 

We apply Coulomb's law between points A and B and again between points A and C.

F=kq1q2r2

Between A and B

FAB=kqAqBrAB2

  • k = 8.99 × 109Nm2/C2 (Coulomb's constant)
  • qA = 1.0 × 10-9 C
  • qB = - 1.0 × 10-9 C
  • rAB = 1 × 10-3 m

Substituting:

FAB=(8.99×109Nm2/C2)(1.0×10-9C)(-1.0×10-9C)(1.0×10-3m)2=(8.99×109Nm2/C2)[(1.0×10-9)(-1.0×10-9)]C21.0×10-6m2

FAB =  - 8.99 × 10-3 N

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Problem Details

A. What is the magnitude of the electric force on charge A in the figure?

F=______N

B. What is the direction of the electric force on charge A in the figure? Choose the best answer.

(a) to the left

(b) to the right

(c) the force is zero