An electric dipole is formed by two charges + q and - q located in...

Electric potential at any general point for vertical dipole is written as, ${V}_{1} ∝ \dfrac{{p}_{1}cos{\theta}_{1}}{{\left({r}_{1}\right)}^{2}}$.

Electric potential at any general point for ${V}_{2} ∝ \dfrac{{p}_{2}cos{\theta}_{2}}{{\left({r}_{2}\right)}^{2}}$.

For first case:

$tan{\theta}_{1} = \dfrac{100}{100} = 1 ⇒ {\theta}_{1} = 45 °$

For second case:

slope in this is $= \dfrac{- 2 - 2}{1 - \left(- 1\right)} = - 2$

which is equal to

$\tan \left(45 ° + {\theta}_{2}\right) = - 2 ⇒ \dfrac{1 + tan{\theta}_{2}}{1 - tan{\theta}_{2}} = - 2 ⇒ 1 + tan{\theta}_{2} = - 2 + 2tan{\theta}_{2} ⇒ tan{\theta}_{2} = 3cos{\theta}_{2} = \dfrac{1}{\sqrt{{3}^{2} + {1}^{2}}} = \dfrac{1}{\sqrt{10}}$

also ${r}_{1} = {r}_{2}$.

Therefore, required ratio

$\dfrac{{V}_{2}}{{V}_{1}} = \dfrac{{p}_{2}cos{\theta}_{2}}{{p}_{1}cos{\theta}_{1}}$

$\dfrac{{V}_{2}}{{V}_{1}} = \dfrac{q{l}_{2} \times \dfrac{1}{\sqrt{10}}}{q{l}_{1} \times \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{{\left(2 - \left(- 2\right)\right)}^{2} + {\left(- 1 - 1\right)}^{2}}}{4 \times \sqrt{5}} = \dfrac{2\sqrt{5}}{4 \times \sqrt{5}} = \dfrac{1}{2}$

$⇒ {V}_{2} = \dfrac{{V}_{0}}{2}$