Question 784: Electrostats & Capacitance

Question

Three point charges $+Q,+2 Q$ and $q$ are placed at the vertices of an equilateral triangle. The value of charge $q$ in terms of $Q$, so that electrical potential energy of the system is zero, is given by

Accepted Answer

Let the three charges $+Q$, $+2Q$, and $q$ be placed at the vertices of an equilateral triangle of side $a$.
Step 1: Potential Energy for a System of 3 Charges
Total electrostatic potential energy, $U$, for three charges $q_1, q_2, q_3$ at the corners of an equilateral triangle (each side = $a$) is:
$$ U = \frac{1}{4\pi \epsilon_0} \left[\frac{q_1 q_2}{a} + \frac{q_2 q_3}{a} + \frac{q_3 q_1}{a} ight] $$
Step 2: Substitute Given Values
Given: $q_1 = Q$, $q_2 = 2Q$, $q_3 = q$.
So,
$ U = \frac{1}{4\pi \epsilon_0} \left[ \frac{Q \cdot 2Q}{a} + \frac{2Q \cdot q}{a} + \frac{q \cdot Q}{a} ight] $
$ U = \frac{1}{4\pi \epsilon_0} \left[\frac{2Q^2}{a} + \frac{2Qq}{a} + \frac{Qq}{a}ight] $
$ U = \frac{1}{4\pi \epsilon_0} \left[\frac{2Q^2 + 3Qq}{a}ight] $
Step 3: Set Potential Energy to Zero
For $U = 0$:
$ 2Q^2 + 3Qq = 0 $
$ 3Qq = -2Q^2 $
$ q = -\frac{2Q^2}{3Q} $
$ q = -\frac{2}{3}Q $
Step 4: Select the Correct Option
The correct answer is:
$ \boxed{	ext{Option C: } q = -\frac{2}{3}Q} $

Answer

$\mathrm{q}=\frac{1}{3} \mathrm{Q}$

Answer

$\mathrm{q}=\frac{2}{3} \mathrm{Q}$

Answer

$\mathrm{q}=-\frac{2}{3} \mathrm{Q}$

Answer

$\mathrm{q}=-\frac{3}{2} \mathrm{Q}$

Electrostats & Capacitance

Three point charges +Q,+2 Q and q are placed at the vertices of an equilateral triangle. The value of charge q in terms of Q, so that electrical potential energy of the system is zero, is given by

Choose the correct answer: