Question

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

Answer 1

Given that,

Three charges +Q, +q and +q are placed at the vertices of a right-angle isosceles triangle.

To find,

If the net electrostatic energy of the configuration is equal to 0, then the value of Q is?

Solution,

Let U₁ is the electrostatic energy between +q (B) and +q (C). So,

$U_1=\dfrac{1}{4\pi \epsilon_o}\dfrac{q^2}{r}$ .....(1)

If U₂ is the electrostatic energy between +Q (A) and +q (B). So,

$U_2=\dfrac{1}{4\pi \epsilon_o}\dfrac{Qq}{r}$ .....(2)

If U₃ is the electrostatic energy between Q (A) and q (C). So,

$U_3=\dfrac{1}{4\pi \epsilon_o}\dfrac{Qq}{\sqrt 2r}$ .....(3)

Net electrostatic energy is :

$U_1+U_2+U_3=0\\\\\dfrac{1}{4\pi \epsilon_o}\dfrac{q}{r}(q+Q+\dfrac{Q}{\sqrt 2})=0\\\\q+Q+\dfrac{Q}{\sqrt 2}=0\\\\-q=Q(1+\dfrac{1}{\sqrt 2})\\\\-q=Q(\dfrac{\sqrt 2+1}{\sqrt 2})\\\\Q=\dfrac{-q\sqrt 2}{\sqrt 2+1}\\\\Q=\dfrac{-2q}{2+\sqrt 2}$

So, the value of Q is equal to $\dfrac{-2q}{2+\sqrt 2}$.

Answer 2

Given that,

Three charges Q, +q and +q  are placed at the vertices of a right angle isosceles triangle.

We know that,

The potential energy of the system

$U=k\dfrac{Q_{1}Q_{2}}{r}$

The net electrostatic energy of the configuration is zero.

We need to calculate the value of Q

For all charges,

$U=U_{q}+U_{q}+U_{q}=0$

$U=k\dfrac{Qq}{a}+\dfrac{kq^2}{a}+\dfrac{kqQ}{a\sqrt{2}}=0$

$\dfrac{Qq}{a}+\dfrac{q^2}{a}+\dfrac{qQ}{a\sqrt{2}}=0$

$\dfrac{Q(\sqrt{2}+1)}{\sqrt{2}}+q=0$

$Q(\sqrt{2}+1)+\sqrt{2}q=0$

$Q=\dfrac{-\sqrt{2}q}{\sqrt{2}+1}$

Hence, The value of Q is $\dfrac{-\sqrt{2}q}{\sqrt{2}+1}$