Question

A glass rod with a net charge of 0.748 μC and a small square of silk cloth with a net charge of −0.502 μC, are separated by 12.0 cm.
What is the magnitude of the attractive force between the two charged objects? You may approximate both objects as point charges.

Answer 1

Answer:

Two positive charge  will produce force of of 0.23 N and negative sign shows that the force between charges is attractive.

Explanation:

Given that,

Net charge on a glass rod, $q_1=0.748\ \mu C=0.748\times 10^{-6}\ C$

Net charge on a silk cloth, $q_1=-0.502\ \mu C=-0.502\times 10^{-6}\ C$

Separation between charges, d = 12 cm = 0.12 m

To find,

The magnitude of the attractive force between the two charged objects.

Solution,

If F is the electric force between two charges. It is given by the electrostatic force of attraction. It is given by :

$F=\dfrac{kq_1q_2}{d^2}$

$F=\dfrac{9\times 10^9\times 0.748\times 10^{-6}\times -0.502\times 10^{-6}}{(0.12)^2}$

$F=-0.23 N$

So, two positive charge  will produce force of of 0.23 N and negative sign shows that the force between charges is attractive.

Answer 2

Answer:

0.235 N

Explanation:

q1 = 0.748 micro Coulomb = 0.748 x 10^-6 C

q2 = - 0.502 micro Coulomb = - 0.502 x 10^-6 C

distance, d = 12 cm = 0.12 m

According to the coulomb's law, the force between the two charged particles is given by

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

$F=\frac{9\times 10^{9}\times 0.748\times 10^{-6}\times 0.502 \times 10^{-6}}{0.12^{2}}$

F = 0.235 N

Thus, the force between the two charges is 0.235 N.