Get Answers to all your Questions

header-bg qa

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to 

  • Option 1)

    \frac{-q}{1 +\sqrt2}

  • Option 2)

    \frac{-\sqrt2 q}{1 +\sqrt2}

  • Option 3)

    -2q

  • Option 4)

    +q

 

Answers (1)

As we have learnt,

 

Potential Energy Of System Of two Charge -

U=\frac{kQ_{1}Q_{2}}{r}  \left ( S.I \right )

U= \frac{Q_{1}Q_{2}}{r}  \left ( C.G.S \right ) 

 

- wherein

K=\frac{1}{4\pi \epsilon _{0}}

 

 Potential energy of the configuration is

U = k\cdot\frac{Qq}{a} + k\frac{q^2}{a} + k\frac{qQ}{a\sqrt2} = 0 \Rightarrow Q = \frac{-\sqrt2 q}{\sqrt2 + 1}

 


Option 1)

\frac{-q}{1 +\sqrt2}

Option 2)

\frac{-\sqrt2 q}{1 +\sqrt2}

Option 3)

-2q

Option 4)

+q

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

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