Question

Two 1.0 kg masses are 1.0 m apart (center to center) on a frictionless table. Each has +10 mC of charge.

a. What is the magnitude of the electric force on one of the masses?

b. What is the initial acceleration of this mass if it is released and allowed to move?

Short answer

a. The magnitude of the electric force on one of the masses is $F_e=0.90N$

b. The initial acceleration of this mass if it is released and allowed to move is $a=0.9m/s^2$

Step-by-step solution

Content Introduction

We are given the information from where we will find the magnitude of the electric force on one of the masses and the initial acceleration of this mass if it is released and allowed to move.

Explanation (Part a)

We are given,

$$\begin{gathered} m_1=1.0kg \\ {m}_2=1.0kg \\ r=1.0m \\ {q}_1=10\mu C \\ {q}_2=10\mu C \end{gathered}$$

Using the formula:

$F_e=\frac{k{q}_1{q}_2}{r^2}$where, K is a Coulomb's constant that is equal to $9\times {10}^{9}N{m}^2/C^2$

Let us first convert the units here,

$\frac{10\mu C}{1}\times \frac{{10}^{-6}C}{1\mu C}=1\times {10}^{-5}C$

Using the above mentioned equation,

$$\begin{gathered} F_e=\frac{k{q}_1{q}_2}{r^2} \\ {F}_e=\frac{(9.0\times {10}^{9}N{m}^2/C^2(1.0\times {10}^{-5}C\left)\right(1.0\times {10}^{-5}C)}{{(1.0m)}^2} \\ {F}_e=0.90N \end{gathered}$$

Explanation (Part b)

To find acceleration, use formula $F=m\times a$

from here we can say,

$a=\frac{F}{m}$where, a is acceleration, F is force and m is mass

$$\begin{gathered} a=\frac{0.9N}{1kg} \\ a=0.9m/s^2 \end{gathered}$$