Question

Two tiny conducting spheres are identical and carry charges of -18.8 µC and +46.5 µC. They are separated by a distance of 2.47 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? N

Answer 1

`F=-12896N`, attractive.

Step-by-step explanation:

For calculating this force we use the Coulomb Law:

`F=(kq_1q_1)/(r^2)`

Where
`k=9*10^9Nm^2/C^(-2)` is the Coulomb's constant,
`q_1` and
`q_2` the values of each charge and r the distance between them.

Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:

`F=((9*10^9Nm^2/C^(-2))(-18.8*10^(-6)C)(46.5*10^(-6)C))/((0.0247m)^2)=-12896N`

This force is attractive since both charges are of opposite sign.