Question

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

• A. ${\displaystyle \frac{-q}{1+\sqrt{2}}}$
• B. $\frac{-2q}{2+\sqrt{2}}$
• C. $-2q$
• D. $+q$

Answer 1

The correct option is B

$\frac{-2q}{2+\sqrt{2}}$

Step 1. Given Data,

Net potential energy is $0$.
$U=0$

Step 2. Formula Used,

Applying the Pythagoras theorem to determine the distance between the charges +$Q$ and $+q$.
$\left(c\right)=\sqrt{{\left(a\right)}^2+{\left(b\right)}^2}$

Potential energy, $U=K{\displaystyle \frac{q_1{q}_2}{r}}$where $$K={\displaystyle \frac{1}{4\pi \epsilon }}$$

The position vector of the positive charge = $r$

$q_1$ and $q_2$ are a group of point charges

Step 3. Calculating the charge $Q$,

Applying the Pythagoras theorem to determine the distance between the charges +$Q$ and$+q$.
$\left(c\right)=\sqrt{{\left(a\right)}^2+{\left(b\right)}^2}$
Substituting the values of ‘$a$’ in the equation to determine the distance between the charges +$Q$ and $+q$.
$$\begin{gathered} \left(c\right)=\sqrt{{\left(a\right)}^2+{\left(b\right)}^2} \\ \Rightarrow \left(c\right)=\sqrt{{\left(a\right)}^2+{\left(a\right)}^2}(a=b) \\ \Rightarrow \left(c\right)=\sqrt{2a^2} \\ \Rightarrow \left(c\right)=\sqrt{2}a \end{gathered}$$
Potential energy $U=K{\displaystyle \frac{q_1{q}_2}{r}}$, $$K={\displaystyle \frac{1}{4\pi \epsilon }}$$.
Net potential energy of the system is given by $U=K{\displaystyle \frac{Qq}{a}}+K{\displaystyle \frac{qq}{a}}+K{\displaystyle \frac{qQ}{a\sqrt{2}}}$

Net potential energy is $0$. So,
$$\begin{gathered} U=0 \\ \Rightarrow K{\displaystyle \frac{Qq}{a}}+K{\displaystyle \frac{qq}{a}}+K{\displaystyle \frac{qQ}{a\sqrt{2}}}=0 \end{gathered}$$
Now solve the equation for $Q$,
$$\begin{gathered} {\displaystyle \frac{Kq}{a}}\left(Q+q+{\displaystyle \frac{Q}{2}}\right)=0 \\ \Rightarrow Q+{\displaystyle \frac{Q}{\sqrt{2}}}=-q \\ \Rightarrow {\displaystyle \frac{Q\sqrt{2}+Q}{\sqrt{2}}}=-q \\ \Rightarrow Q\left(1+\sqrt{2}\right)=-q\sqrt{2} \\ \therefore Q={\displaystyle \frac{-q\sqrt{2}}{\left(1+\sqrt{2}\right)}} \end{gathered}$$

$\Rightarrow Q=\frac{-2q}{2+\sqrt{2}}$
Therefore, the value of the charge “$Q$” such that the net electronic energy of the configuration of the charges shown in the figure is equivalent to zero is$\frac{-2q}{2+\sqrt{2}}$.
Hence, option B is correct.