Get Answers to all your Questions

header-bg qa

Four equal point charges Q each are placed in the xy plane at (0,2), (4,2), (4, -2) and (0, -2). The work required to put a fifth charge Q at the ogigin of the coordinate system will be :

 

  • Option 1)

     

    \frac{Q^{2}}{4\pi \varepsilon _{0}}\left ( 1+\frac{1}{\sqrt{3}} \right )

  • Option 2)

     

    \frac{Q^{2}}{4\pi \varepsilon _{0}}\left ( 1+\frac{1}{\sqrt{5}} \right )

  • Option 3)

     

    \frac{Q^{2}}{4\pi \varepsilon _{0}}

  • Option 4)

    \frac{Q^{2}}{2\sqrt{2}\pi \varepsilon _{0}}

Answers (1)

best_answer

 

Potential Energy Of a System Of n Charge -

U= K\left ( \frac{Q_{1}Q_{2}}{r_{12}} +\frac{Q_{2}Q_{3}}{r_{23}}+\frac{Q_{1}Q_{3}}{r_{13}}\right )

- wherein

For system of 3 charges.

 

Vorigin = 2\frac{KQ}{2}+2\frac{KQ}{\sqrt{20}}

           = KQ\left ( 1+\frac{1}{\sqrt{5}} \right )

\therefore Work required to put a 5th charge Q at origin

= Q Vorigin

=\frac{Q^{2}}{4\pi \epsilon _{0}} \left [ 1+\frac{1}{\sqrt{5}} \right ]


Option 1)

 

\frac{Q^{2}}{4\pi \varepsilon _{0}}\left ( 1+\frac{1}{\sqrt{3}} \right )

Option 2)

 

\frac{Q^{2}}{4\pi \varepsilon _{0}}\left ( 1+\frac{1}{\sqrt{5}} \right )

Option 3)

 

\frac{Q^{2}}{4\pi \varepsilon _{0}}

Option 4)

\frac{Q^{2}}{2\sqrt{2}\pi \varepsilon _{0}}

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE