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An arc of radius r carries charge. The linear density of charge is  \lambda and the arc subtends a angle \frac{\pi }{3}   at the centre. What is electric potential at the centre 

  • Option 1)

    \frac{\lambda }{4\varepsilon _{0}}

     

     

     

  • Option 2)

    \frac{\lambda }{8\varepsilon _{0}}

  • Option 3)

    \frac{\lambda }{12\varepsilon _{0}}

  • Option 4)

    \frac{\lambda }{16\varepsilon _{0}}

 

Answers (2)

As we learned

 

Electric Potential due to Continious charge distribution -

\dpi{100} V=\int dV=\int \frac{dq}{4\pi \varepsilon _{0}r}

-

 

 

Length of the arc =r\Theta =\frac{r\pi }{3}

Charge on the arc  =\frac{r\pi }{3}\times \lambda

\ Potential at center  = \frac{kq}{r}

=\frac{1}{4\pi \varepsilon _{0}}\times \frac{r\pi \lambda }{3}\frac{\lambda }{r}=\frac{\lambda }{12\varepsilon _{0}}


Option 1)

\frac{\lambda }{4\varepsilon _{0}}

 

 

 

Option 2)

\frac{\lambda }{8\varepsilon _{0}}

Option 3)

\frac{\lambda }{12\varepsilon _{0}}

Option 4)

\frac{\lambda }{16\varepsilon _{0}}

Posted by

Vakul

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option 3

 

Posted by

Darshan jain

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