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A charge particle q is shifted from point 1 to 2 in electric feild due to staright long linear charge of \lambda. find the potential difference between 1 and 2.

  • Option 1)

    0

  • Option 2)

    \infty

  • Option 3)

    \frac{\lambda}{2\pi\epsilon_0}\ln 2

  • Option 4)

    \frac{\lambda}{2\pi\epsilon_0 r}

 

Answers (1)

best_answer

As we have learnt,

 

Potential due to line charge -

V=\frac{\lambda }{2\pi \varepsilon _{0}}\log_{e}\left [ \frac{\sqrt{r^{2}+l^{2}}-l}{\sqrt{r^{2}+l^{2}}+l} \right ]

-

 

 \Delta V = \frac{W}{q}\Rightarrow w = \int F\cdot dr = q\int E\cdot dr = q\int \frac{\lambda}{2\pi\epsilon_0 r}\cdot dr \\*= \frac{\lambda q}{2\pi\epsilon_0}\int_{R}^{2R}\frac{dr}{r} = \frac{\lambda q}{2\pi\epsilon_0}\ln 2

\Delta V = \frac{\frac{\lambda q}{2\pi\epsilon_0}\ln 2}{q} = {\frac{\lambda }{2\pi\epsilon_0}\ln 2}

 


Option 1)

0

Option 2)

\infty

Option 3)

\frac{\lambda}{2\pi\epsilon_0}\ln 2

Option 4)

\frac{\lambda}{2\pi\epsilon_0 r}

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