Three charges Q-q-q are placed the vertices of an equilateral triangle- If the net electric potential energy of the system is 0- then Q is-qq0None of the above

Question

Toppr Admin · Accepted Answer

U-UAB-UAC-UBC-x27F9-U-14-x3C0-x3F5-o-qQl-x2212-qQl-x2212-q-ql-0-x27F9-q-Q-x2212-Q-x2212-q-0This gives both q-Q-0And- for any vlaueQ- potential energy of this system can-apos-t be 0-

Three charges $Q, q, - q$ are placed at the vertices of an equilateral triangle. If the net electric potential energy of the system is $0$ , then $Q$ is
$- q$
$q$
$0$
None of the above

$U = U_{A B} + U_{A C} + U_{B C}$

$⟹ U = \frac{1}{4 π ϵ_{o}} (\frac{q Q}{l} + \frac{- q Q}{l} + \frac{- q . q}{l}) = 0$

$⟹ q (Q - Q - q) = 0$

This gives both $q = Q = 0$

And, for any vlaue $Q$ , potential energy of this system can't be $0$ .

Three charges Q,q,−q are placed at the vertices of an equilateral triangle. If the net electric potential energy of the system is 0, then Q is−qq0None of the above

U=UAB+UAC+UBC⟹U=14πϵo(qQl+−qQl+−q.ql)=0⟹q(Q−Q−q)=0This gives both q=Q=0And, for any vlaueQ, potential energy of this system can't be 0.

Three charges $Q, q, - q$ are placed at the vertices of an equilateral triangle. If the net electric potential energy of the system is $0$ , then $Q$ is
$- q$
$q$
$0$
None of the above

$U = U_{A B} + U_{A C} + U_{B C}$

$⟹ U = \frac{1}{4 π ϵ_{o}} (\frac{q Q}{l} + \frac{- q Q}{l} + \frac{- q . q}{l}) = 0$

$⟹ q (Q - Q - q) = 0$

This gives both $q = Q = 0$

And, for any vlaue $Q$ , potential energy of this system can't be $0$ .