Question

Two particles, one with charge -5.45 uC and one with charge 3.55 uC, are 4.34 cm apart. What is the magnitude of the force that one particle exerts on the other?

Answer 1

Answer:

Force, F = 92.02 N

Explanation:

It is given that,

Charge 1, $q_1=-5.45\ \mu C=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=3.55\ \mu C=3.55\times 10^{-6}\ C$

Distance between two charges, d = 4.35 cm = 0.0435 m

We need to find the magnitude of the force that one particle exerts on the other. It is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\ C\times 3.55\times 10^{-6}\ C}{(0.0435\ m)^2}$

F = -92.02 N

So, the force acting between two particles is 92.02 N. Hence, this is the required solution.

Answer 2

Answer:

$F = 92.45 N$

Explanation:

As we know that the force between two charge particles is given by

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

here we know that

$q_1 = 3.55 \mu C$

$q_2 = 5.45 \mu C$

now the distance between the two charges is

r = 4.34 cm

now from the formula of electrostatic force we will have

$F = \frac{(9\times 10^9)(3.55 \mu C)(5.45 \mu C)}{0.0434^2}$

$F = 92.45 N$