Answer:
The electric potential will be "259.695 volt".
Explanation:
In the given question, the figure is not provided. Below is the attached figure given.
Given:
[tex]q_1=6.39\times 10^{-9} \ C[/tex]
[tex]q_2=3.22\times 10^{-9} \ C[/tex]
[tex]AP=(0.150+0.250)[/tex]
[tex]=0.40 \ m[/tex]
[tex]BP=0.25 \ m[/tex]
Now,
At point P, the electric potential will be:
⇒ [tex]V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}[/tex]
By putting values, we get
⇒ [tex]=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ][/tex]
⇒ [tex]=259.695 \ Volt[/tex]
