Three charges, \( Q \), \( -q \) and \( 2q \) are placed at the vertices of a right-angled isosceles triangle. What is the value of \( q \) for the net electrostatic energy of the configuration to be zero?
Show Hint
For electrostatic energy of a system of charges, calculate pairwise potential energies and add them algebraically with signs of charges.
Step 1: Identify the distances between charges.
The triangle is a right-angled isosceles triangle.
The two perpendicular sides are each \( r \).
So, the distance between \( Q \) and \( 2q \) is:
\[
r
\]
The distance between \( 2q \) and \( -q \) is:
\[
r
\]
The distance between \( Q \) and \( -q \) is the hypotenuse:
\[
\sqrt{r^2+r^2}=r\sqrt{2}
\] Step 2: Write electrostatic potential energy formula.
For two charges \( q_1 \) and \( q_2 \) separated by distance \( r \):
\[
U = \frac{kq_1q_2}{r}
\]
Total electrostatic energy is the sum of energies of all pairs of charges. Step 3: Write energy between \( Q \) and \( 2q \).
\[
U_1 = \frac{k(Q)(2q)}{r}
\]
\[
U_1 = \frac{2kQq}{r}
\] Step 4: Write energy between \( 2q \) and \( -q \).
\[
U_2 = \frac{k(2q)(-q)}{r}
\]
\[
U_2 = -\frac{2kq^2}{r}
\] Step 5: Write energy between \( Q \) and \( -q \).
The distance between \( Q \) and \( -q \) is \( r\sqrt{2} \).
\[
U_3 = \frac{k(Q)(-q)}{r\sqrt{2}}
\]
\[
U_3 = -\frac{kQq}{r\sqrt{2}}
\] Step 6: Apply condition for zero net electrostatic energy.
\[
U_1+U_2+U_3=0
\]
\[
\frac{2kQq}{r}-\frac{2kq^2}{r}-\frac{kQq}{r\sqrt{2}}=0
\]
Taking \( \frac{kq}{r} \) common:
\[
\frac{kq}{r}\left(2Q-2q-\frac{Q}{\sqrt{2}}\right)=0
\]
Since \( \frac{kq}{r}\neq 0 \), we get:
\[
2Q-2q-\frac{Q}{\sqrt{2}}=0
\] Step 7: Solve for \( q \).
\[
2q = 2Q-\frac{Q}{\sqrt{2}}
\]
\[
q = Q-\frac{Q}{2\sqrt{2}}
\]
\[
q = Q\left[1-\frac{1}{2\sqrt{2}}\right]
\]
Therefore:
\[
\boxed{Q\left[1-\frac{1}{2\sqrt{2}}\right]}
\]
