Question

A test charge of +2μC is placed halfway between a charge of +6μC and another of +4μC separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6μC charge)?

Answer 1

Answer:

Part a)

F = 14.4 N

Part b)

away from +6μC charge

Explanation:

Force on the charge placed midway between two charges is given by

$F = \frac{kq_1q}{r^2} - \frac{kq_2q}{r^2}$

here we know that

$q = 2\mu C$

$q_1 = 6 \mu C$

$q_2 = 4\mu C$

here the charge is placed at mid point so we have

$r = 5 cm$

now we have

$F = \frac{(9\times 10^9)(6\mu C)(2\mu C)}{0.05^2} - \frac{(9\times 10^9)(2\mu C)(4\mu C)}{0.05^2}$

$F = 43.2 - 28.8 = 14.4 N$

Part b)

since +6μC charge is more in magnitude so the force due to this charge will be more so the net force on it is away from +6μC charge