Question

Object A, which has been charged to +14.83 nC, is at the origin. Object B, which has been charged to -25.38 nC, is at x=0 and y=2.55 cm. What is the magnitude of the electric force, in micro-Newtons, on object A?

Answer 1

Given:

Object A has the charge,

$\begin{gathered} +14.83\text{ nC} \\ =+14.83\times10^{-9}\text{ C} \end{gathered}$

This charge is at the origin.

Object B has the charge,

$\begin{gathered} -25.38\text{ nC} \\ =-25.38\times10^{-9}\text{ C} \end{gathered}$

Object B is at the point,

$\begin{gathered} (x,\text{ y\rparen=\lparen0, 2.55 cm\rparen} \\ =(0,\text{ 0.0255 m\rparen} \end{gathered}$

To find:

The magnitude of the electric force, in micro-Newtons, on object A

Explanation:

The electric force between two charges is given by,

$\begin{gathered} F=k\frac{q_1q_2}{r^2} \\ K=9\times10^9\text{ N.m}^2.C^{-2} \end{gathered}$

The distance between A and B is,

$\begin{gathered} r=\sqrt{0^2+(0.0255)^2} \\ =0.0255\text{ m} \end{gathered}$

The force will be directed towards the negative charge and the magnitude is,

$\begin{gathered} F=9\times10^9\frac{14.83\times10^{-9}\times25.38\times10^{-9}}{(0.0255)^2} \\ =5209.5\times10^{-6}\text{ N} \\ =5209.5\text{ }\mu N \end{gathered}$

Hence, the electric force on A is,

$5209.5\text{ }\mu N$