The charges Q, +q and + q are placed at the vertices of an equilateral triangle of side l . If the net electrostatic potential energy of the system is zero , the Q is equal to

Options

A. `+q//2`
B. zero
C. `-q//2`
D. `-q`

Answer

C

Explanation

Potential energy A due to
`B=(1)/(4piepsilon_(0))*(q^(2))/(a)`
Potential energy of B due to
`C=(1)/(4piepsilon_(0))*(Qq)/(q)`.
Potential energy of C due to `A=(1)/(4piepsilon_(0))*(Qq)/(a)`.
`therefore` Total potential energy `=(1)/(4piepsilon_(0))*(q^(2))/(a)+(2Qq)/(4piepsilon_(0)a)=0`.
Therefore Q has to be negative .
`q^(2)=-2QqrArrQ=-q//2`.