What is the force on the 1.0nC charge in the middle of the figure due to the four other charges?
F = K*q1*q2 / r^2
K = 8.99x10^9
The Attempt at a Solution
Initial observation: The symmetry of the charges means that Fy = 0 so I only need to worry about Fx.
F of P1 on A = - K*q[P1]*q[A] / r^2
r^2 = a^2 + b^2 = 0.5^2 + 0.5^2 = 0.25 + 0.25 = 0.5
K*q[P1]*q[A] / 0.5 = - K*q[P1]*q[A]*2
Second observation: The force of P1 on A is equal to the force of P2 on A
Therefore the force of both P1 and P2 on A is
Third observation: The force of N1 and N2 on A is equal to the force of P1 and P2 on A
Therefore the force of N1, N2, P1 and P2 on A (which is Fx) is
- K*q[P1]*q[A]*2*2*2 = - K*q[P1]*q[A]*8
- (8.99x10^9)(1x10^-19)(2x10^-19)*8 = 0.0000000000000000000000000014384
Where did I go wrong? Thanks in advance!
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