Question 47: Electrostatics & Capacitance

Question

Three charges, $Q,-q$ and $2 q$ 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?

Accepted Answer

The kinetic energy gained is equal to the work done by the field:
$$K=qEd$$
Here
Ã¢ÂÂ¢ $q=60	imes10^{-6}$ C
Ã¢ÂÂ¢ $E=10^5$ N/C
Ã¢ÂÂ¢ $d=0.02$ m
So
$$K=(60	imes10^{-6})(10^5)(0.02)=0.12\ {m J}=12	imes10^{-2}\ {m J}$$
Hence $m=12$. (Option A)

Answer

$Q\left[\frac{1}{2 \sqrt{2}}-1\right]$

Answer

$Q\left[1-\frac{1}{2 \sqrt{2}}\right]$

Answer

$Q\left[1+\frac{1}{2 \sqrt{2}}\right]$

Answer

$Q\left[\frac{2 \sqrt{2}+1}{2}\right]$

Electrostatics & Capacitance

Three charges, Q,-q and 2 q 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?

Choose the correct answer: