Question

Object A, which has been charged to +4.0 nC, is at the origin. Object B, which has been charged to -8.0 nC, is at 1x, y2 = 10.0 cm, 2.0 cm2. Determine the electric force on each object. Write each force vector in component form

Short answer

${\stackrel{\rightharpoonup }{F}}_{4nCon8nC}=-7.2\times {10}^{-4}jN$and${\stackrel{\rightharpoonup }{F}}_{8nCon4nC}=7.2\times {10}^{-4}jN$

Step-by-step solution

Content Introduction

We have two charged particles one is$q_1$and the other is$q_2$with the distance$r$ the two particles exert forces on each other where the force decrease with the increase in the distance and increase with the increase in magnitude.

Content Explanation

Coulomb's law describe

$F_C=\frac{k{q}_1{q}_2}{r^2}$where K is electrostatic constant by $9.0\times {10}^{9}N{m}^2/C^2$.

The magnitude of the electric force is $+4.0nCon-8nC$

$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_{4nCon8nC}=\frac{K{q}_1{q}_1}{r^2} \\ {\stackrel{\rightharpoonup }{F}}_{4nCon8nC}=\frac{(9.0\times {10}^{9}N{m}^2/C^2)(+4\times {10}^{-9}C)(-8\times {10}^{-9}C)}{{(0.02m)}^2}\left(j\right) \\ {\stackrel{\rightharpoonup }{F}}_{4nCon8nC}=-7.2\times {10}^{-4}\overbrace{j}N \end{gathered}$$

The magnitude of the electric force is $-8nCon+4nC$

$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_{8nCon4nC}=\frac{K{q}_1{q}_1}{r^2} \\ {\stackrel{\rightharpoonup }{F}}_{8nCon4nC}=\frac{(9.0\times {10}^{9}N{m}^2/C^2)(-8\times {10}^{-9}C)(+4\times {10}^{-9}C)}{{(0.02m)}^2}\left(j\right) \\ {\stackrel{\rightharpoonup }{F}}_{8nCon4nC}=7.2\times {10}^{-4}\overbrace{j}N \end{gathered}$$