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

0.235 N

Step-by-step 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=(Kq_(1)q_(2))/(d^(2))`

`F=(9* 10^(9)* 0.748* 10^(-6)* 0.502 * 10^(-6))/(0.12^(2))`

F = 0.235 N

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

Answer 2

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

Step-by-step explanation:

Given that,

Net charge on a glass rod,
`q_1=0.748\ \mu C=0.748* 10^(-6)\ C`

Net charge on a silk cloth,
`q_1=-0.502\ \mu C=-0.502* 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=(kq_1q_2)/(d^2)`

`F=(9* 10^9* 0.748* 10^(-6)* -0.502* 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.