Question

A negative charge -0.550 HC exerts an upward 0.700 N force on an unknown charge 0.220 m directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the -0.550 HC charge? downward upward

Answer 1

Answer:

Part a)

$q_2 = -6.8 \mu C$

Part b)

$F = 0.700 N$

direction = downwards

Explanation:

As we know that the negative charge will experience the force due to some other charge below it

the force is given as

$F = 0.700 N$

now we know that

$F = \frac{kq_1q_2}{r^2}$

now plug in all data

$0.700 = \frac{(9 \times 10^9)(0.550\mu C)q_2}{0.220^2}$

$0.700 = 1.022\times 10^5 q_2$

$q_2 = -6.8 \mu C$

since this is a repulsion force so it must be a negative charge

Part b)

As per Newton's III law it will exert equal and opposite force on it

So here the force on the charge below it will be same in magnitude but opposite in direction

so here we have

$F = 0.700 N$

direction = downwards