Two point charges are placed on the x-axis as follows: charge q1 = 3.98 nC is located at x = 0.196 m, and charge q2 = 4.97 nC is at x =-0.296 m.

Question

Question

Part A

What is the magnitude of the total force exerted by these two charges on a negative point charge q3 =-6.03 nC that is placed at the origin?

Part B

What is the direction of the total force exerted by these two charges on a negative point charge q3 = -6.03 nC that is placed at the origin?

Answer 1

Answer #1

Given that q1= 3.98*10^-9 C , x1 = 0.196 m
q2 = 4.97*10^-9 x2 = -0.296 m

Here ,

electric force between two charges is given as

F = k*q1*q2/d^2

Now, for the charge placed at origin

Fnet = F1 - F2

Fnet = 9*10^9 * (0.603 *10^-9 * 10^-9) *(3.98/.196^2 i - i 4.97/.296^2)

solving for Fnet

Fnet = 2.605 *10^-7 i N

the net force on charge at the origin is 2.605 *10^-7 N

b) the net force is in +ve x - direction

Answer 2

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