JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Electrostatics

The fields of two unlike charges cancel only outside the pair, beyond the weaker charge. Let the test charge sit a distance $y$ from the $-5\mu \mathrm{C}$, making it $5+y$ (in cm) from the $20\mu \mathrm{C}$. Equating field magnitudes, $\frac{20}{(5+y{)}^2}=\frac{5}{y^2}$ gives $\frac{4}{(5+y{)}^2}=\frac{1}{y^2}$, so $2y=5+y$ and $y=5$ cm. The particle must be placed $5$ cm from the $-5\mu \mathrm{C}$ charge, on the side away from the $20\mu \mathrm{C}$ charge.