What is the net electric force acting on Q1?
- 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.
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.
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