# What is the net electric force acting on Q1?

### 2 Answers

- Steve4PhysicsLv 71 month ago
The force on Q1 from Q2 is k.Q1.Q2/d²

= 9*10^9 * 200*10^-6 * 100*10^-6 / 4.0² = 11.25N

This is positive indicating repulsion. Q1 is pushed away from Q2. So the force on Q1 from Q2 is 11.25N to the left.

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The force on Q1 from Q3 is k.Q1.Q3/d²= 9*10^9 * 200*10^-6 * (-50)*10^-6 / 6.0² = -2.5N

This is negative indicating attraction. Q1 is pulled towards Q3. So the force on Q1 from Q3 is 2.5N to the right.

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Net force = 11.25 - 2.5 = 8.75N to the left.

If you are using (+x = right) and (-x = left) then the forces are -11.25N and 2.5N. The net force is -11.25 + 2.5 = -8.75N

- billrussell42Lv 71 month ago
force due to Q2:

F2 = (9e9)(200e-6)(100e-6) / (4)² = 180000e-3/16 = 180/16 = 11.25 N, to the left

force due to Q3:F3 = (9e9)(200e-6)(50e-6) / (6)² = 90000e-3/36 = 90/16 = 5.63 N, to the right

net is 5.63 N to the left

Coulomb's law, force of attraction/repulsion

F = kQ₁Q₂/r²

Q₁ and Q₂ are the charges in coulombs

F is force in newtons

r is separation in meters

k = 8.99e9 Nm²/C²