Question

Three origin, the identical second point to the charges right of atx 2.0 hc 50 cm, are and placed the on thrd the 1s at x-axis. the 100 the cm first mark. charge 15 what at are the the magnitude le and direction of the electrostatic torce which acts on the charge at the or ongin?

Answer 1

The magnitude and direction of the electrostatic torce is - 0.00712 x 10^9N towards left direction

Given:

q1 = +2 uC

q2 = -2 uC

q3 = +4 uC

To Find:

magnitude le and direction of the electrostatic torce

Solution: Electric force is a vector quantity. The electric force is proportional to the product of the magnitude of the charges. It is inversely proportional to the square of the distance between the charges. The electric force is calculated by,

F = kqQ/r^2

Force between q1 and q2

F1 = +2 x -2 x k/50 x 50 = - 0.0016k

Force between q1 and q3

F2 = +2 x +4 x k/100 x 100 = + 0.0008k

Net force on charge at origin is F1 + F2

F(net) = F1 + F2 = - 0.0016k + 0.0008k

F(net) = - 0.0008 x 8.9 x 10^9 = - 0.00712 x 10^9N

So, force on charge at origin is - 0.00712 x 10^9N and towards left direction

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