Geobassurseuad5

2023-02-20

A point charge ${q}_{1}=-5.8\mu C$ is held stationary at the origin. A second charge $q2=+4.3\mu C$ moves from the point (0.26 m,0,0) to (0.38 m,0,0). Find the workdone by the electric force on ${q}_{2}$.

A. −0.272 J

B. +0.272 J

C. −0.381 J

D. +0.381 J

A. −0.272 J

B. +0.272 J

C. −0.381 J

D. +0.381 J

mandyllau34fy

Beginner2023-02-21Added 5 answers

Given,

${q}_{1}=-5.8\mu C,{q}_{2}=+4.3\mu C$

${r}_{i}=0.26m,{r}_{f}=0.38m$

Workdone by the electrostatic force $W={U}_{i}-{U}_{f}$

$\because \Delta U=\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon}_{0}}(\frac{1}{{r}_{f}}-\frac{1}{{r}_{i}})$

We can write that,

$W=\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon}_{0}}(\frac{1}{{r}_{i}}-\frac{1}{{r}_{f}})$

By substituting the received data,

$W=\frac{(-5.8\times 10-6)(4.3\times 10-6)(9\times 109)(0.38-0.26)}{0.38\times 0.26}$

$\Rightarrow W=-0.272J$

Therefore, option (a) is the correct answer.

${q}_{1}=-5.8\mu C,{q}_{2}=+4.3\mu C$

${r}_{i}=0.26m,{r}_{f}=0.38m$

Workdone by the electrostatic force $W={U}_{i}-{U}_{f}$

$\because \Delta U=\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon}_{0}}(\frac{1}{{r}_{f}}-\frac{1}{{r}_{i}})$

We can write that,

$W=\frac{{q}_{1}{q}_{2}}{4\pi {\epsilon}_{0}}(\frac{1}{{r}_{i}}-\frac{1}{{r}_{f}})$

By substituting the received data,

$W=\frac{(-5.8\times 10-6)(4.3\times 10-6)(9\times 109)(0.38-0.26)}{0.38\times 0.26}$

$\Rightarrow W=-0.272J$

Therefore, option (a) is the correct answer.