c
\(E _{2}=\) electric field due to \(Q _{2}\)

\(=\frac{ kQ _{2}}{ x _{2}^{2}}\)

\(E _{1}=\frac{ k Q _{1}}{ x _{1}^{2}}\)

From diagram

\(\tan \theta=\frac{ E _{2}}{ E _{1}}=\frac{ x _{1}}{ x _{2}}\)

\(\frac{ kQ _{2}}{ x _{2}^{2} \times \frac{ k Q _{1}}{ x _{1}^{2}}}=\frac{ x _{1}}{ x _{2}}\)

\(\frac{ Q _{2} x _{1}^{2}}{ Q _{1} x _{2}^{2}}=\frac{ x _{1}}{ x _{2}}\)

\(\frac{Q_{2}}{Q_{1}}=\frac{x_{2}}{x_{1}}\)

\(\frac{Q_{1}}{Q_{2}}=\frac{x_{1}}{x_{2}}\)