b
Electric field at \(P\) due \((-q)\) will be in direction \(P B\) given by \(E_1\)

\(E_1=\frac{1}{4 \pi \varepsilon_0}+\frac{q}{\sqrt{x^2+a^2}}(\cos \theta \hat{\imath}+\sin \theta \hat{\jmath})\)

Fectric field at \(p\) due \((+2)\) will be

\(E_2=\frac{1}{4 \pi t_0} \frac{q}{\sqrt{x^2+a^2}}(\cos \hat{\imath}-\sin \theta \hat{\jmath})\)

Total Electric field \(t E=E_1+E_2\)

\(E =\frac{1}{4 \pi \epsilon_0} \frac{2}{\sqrt{x^2+q^2}}\left[2 \cos \theta \hat{\imath}+\sin \theta \hat{\jmath}-\sin \theta e_{ l }\right]\)

\(=\frac{2 q 2 \cos \theta \hat{\imath}}{4 \pi \epsilon_0 \sqrt{x^2+q^2}}\)

Total potential will be \(v=v_1+r_2\)

\(V=\frac{1}{4 \pi t_0} \frac{q}{\sqrt{x^2+a^2}}-\frac{1}{4 \pi t_0} \frac{q}{\sqrt{x^2+a^2}}=0\)