Three 100 nCcharged objects are equally spaced on a straight line. The separation of each object from its neighbor is 0.3 m. Find the force exerted on the center object if the rightmost charge is negative and the other two are positive.

0N

Explanation:

Parameters given:

Q1 = Q2 = Q3 = 100 * 10^(-9) C

r = 0.3 m

The net force on the middle charge is:

F = F(1,2) + F(2,3)

Where F(1,2) = force due to Q1 on Q2

F(2,3) = force due to Q3 on Q2

F(1,2) = k*Q(1)*Q(2)/r²

F(1,2) = [9 * 10^9 * 100 * 10^(-9) * 100 * 10^(-9)]/(0.3²)

F(1,2) = 0.001N

F(2,3) = k*Q(2)*Q(3)/r²

F(2,3) = [9 * 10^9 * 100 * 10^(-9) * -100 * 10^(-9)]/(0.3²)

F(2,3) = -0.001N

Therefore,

F = 0.001 - 0.001

F = 0N

answer would be d.) i hope this

explanation:

double the distance = divide strength by 4