Question

Two 2.0 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge. PART A

Answer 1

Answer:

A) Force = 0.69 N

B) Acceleration = 0.34 m/s^2

Explanation:

The electric force is given by the expression:

$F_e= K *\frac{q_1*q_2}{r^2}$

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and 12 is the charge of the particles, and r is the distance:

$F_e = 9*10^9 Nm^2/C^2 * \frac{(9.6*10^{-6}C)^2}{(1.1m)^2} = 0.69 N$

Part B.

For the acceleration, you need Newton's second Law:

F = m*a

Then,

$a = \frac{F}{m} = \frac{0.69 N}{2 kg} = 0.34 m/s^2$