# Wiz the cat is batting at two table tennis balls hanging from insulating threads with their sides just barely touching. Each ball acquires a positive  charge of `3.5 xx10^(-9)C`  from Wiz’s fur and they swing apart.   a) If a force of `6.0 xx10^(-5)N`  acts on one of the balls, how far apart are they from each other? b) Is the force between them one of attraction or repulsion? During the friction with the fur of the cat, the balls acquire electric charge. As the balls are spherical, we can consider that the charge is situated in its Center and we can apply the Coulomb's law for the electrostatic interaction, between two particles charged.

a)

F = (Kq1q2)/d2

where:

K = 9*10^9 Nm^2/C^2, the electric constant for the vacuum.

q1 = q2, the electric charges that the tennis balls acquire during the rubbing with the cat.

d, is the distance measured between the centers of the tennis balls.

Solving for the distance d and substituting, we have:

d = sqrt(Kq^2)/F

d = sqrt(9*10^9(3.5*10^-9)^2)/(6.0 x 10^-5)

d = 4.28*10^-2 m = 4.28 cm

The distance between the centers of the balls is d = 4.28 cm.

b) The balls acquire electric charges of the same sign, so the force is repulsive.

Approved by eNotes Editorial Team (a) To solve, apply Coulomb's Law.

`F = k (q_1q_2)/r^2`

where k is the electrostatic constant `9 xx 10^9 N*m^2/C^2` , q1 and q2 are the charges in Coulombs and r is the distance in meters between the two charges.

Plugging in the given values, the formula becomes:

`6xx10^(-5) = 9xx10^9 * ((3.5xx10^(-9))(3.5xx10^(-9)))/r^2`

`6xx10^(-5) = (1.1025xx10^(-7))/r^2`

`r^2=(1.1025xx10^(-7))/(6xx10^(-5))`

`r=sqrt((1.1025xx10^(-7))/(6xx10^(-5)))`

`r=0.0429`

Therefore, the two balls are 0.0429 meters away from each other.

(b) Since the two balls have like charges, they will repel each other.

Approved by eNotes Editorial Team