Question

A point charge of 6.30 uC is lecated 15.0 cm from a second point charge of-4.80 yC. The force on the 6.30 juC charge is

Answer 1

Answer:

12.096 N

Explanation:

Q = 6.30 micro coulomb = 6.3 x 10^-6 C

q = - 4.8 micro coulomb = - 4.8 x 10^-6 C

d = 15 cm = 0.15 m

According to the Coulomb's law, the force between the two charges is given by

$F = \frac{KQq}{d^{2}}$

Where, k be the Coulombic constant which is equal to 9 x 10^9 Nm^2 / C^2. This value is for free space or vacuum be the medium between two charges.

By substituting the values

$F = \frac{9\times 10^{9}\times 6.3 \times10^{-6}\times 4.8\times 10^{-6}}{0.15\times 0.15}$

F = 12.096 N

This force is attractive in nature as the charges as unlike in nature and by the properties of charges, if there are two unlike charges, then the force between the charges is attractive in nature.