Total potential at P due to whole

\(\displaystyle{V}={\left(\frac{\sigma}{{2}}\sigma_{{0}}\right)}{\left(\sigma{R}^{{2}}+{D}^{{2}}\right)}-{D}\sigma\)

\(\displaystyle={6.28}{f}\frac{{C}}{{m}^{{2}}}\)

\(\displaystyle={6.28}\cdot{10}^{{-{15}}}\ \frac{{C}}{{m}^{{2}}}\sigma_{{0}}\)

\(\displaystyle={8.85}\cdot{10}^{{-{12}}}{\frac{{{C}^{{2}}}}{{{N}{m}^{{2}}}}}\)

\(\displaystyle{R}={64}{c}{m}={0.64}{m}\)

\(\displaystyle{D}={25.9}{c}{m}={0.259}{m}\)

PLUG THE VALUES WE GET V

After getting V POTENTIAL DUE TO ONE QUARTER \(\displaystyle={\frac{{{V}}}{{{4}}}}\)

\(\displaystyle={1.357}\cdot{10}^{{-{5}}}\)

\(\displaystyle{V}={\left(\frac{\sigma}{{2}}\sigma_{{0}}\right)}{\left(\sigma{R}^{{2}}+{D}^{{2}}\right)}-{D}\sigma\)

\(\displaystyle={6.28}{f}\frac{{C}}{{m}^{{2}}}\)

\(\displaystyle={6.28}\cdot{10}^{{-{15}}}\ \frac{{C}}{{m}^{{2}}}\sigma_{{0}}\)

\(\displaystyle={8.85}\cdot{10}^{{-{12}}}{\frac{{{C}^{{2}}}}{{{N}{m}^{{2}}}}}\)

\(\displaystyle{R}={64}{c}{m}={0.64}{m}\)

\(\displaystyle{D}={25.9}{c}{m}={0.259}{m}\)

PLUG THE VALUES WE GET V

After getting V POTENTIAL DUE TO ONE QUARTER \(\displaystyle={\frac{{{V}}}{{{4}}}}\)

\(\displaystyle={1.357}\cdot{10}^{{-{5}}}\)