Question

Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on the balls are -3.31x10-7 C and -5.7x10-7 C. Determine the force of electrical repulsion in Newtons when held a distance of 22 cm apart. ________N (round to two decimal places) LaTeX: F=k\frac{q^1q^2}{d^2} F = k q 1 q 2 d 2 k = 9.0x109 Nm2/C2

Answer 1

Given Information:  

Charge on ball 1 = q₁ = -3.31x10⁻⁷ C

Charge on ball 2 = q₂ = -5.7x10⁻⁷ C

Distance = r = 22 cm = 0.22 m

Required Information:  

Force of electrical repulsion = F = ?

Answer:

Force of electrical repulsion = 3.51x10⁻² N

Explanation:

The force of electrical repulsion is given by

F = kq₁q₂/r²

Where k is the coulomb constant, q₁ is the charge on ball 1 and q₂ is the charge on ball 2 and r is the distance between these balls.

F = (9x10⁹*-3.31x10⁻⁷*-5.7x10⁻⁷)/(0.22)²

F = 3.51x10⁻² N

Therefore, the force of electrical repulsion between two balls is 3.51x10⁻² N.

Answer 2

Answer:

0.035 N

Explanation:

Parameters given:

Charge q1 = -3.31x10^(-7) C

Charge q2 = -5.7x10^(-7) C.

Distance between them, R = 22 cm = 0.22 m

Electrostatic force between to particles is given as:

F = (k* q1 * q2) / R²

F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²

F = 0.035 N