Question

In the diagram, q1 = +6.39*10^_9 C and

Answer 1

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:

$q_1=6.39\times 10^{-9} \ C$

$q_2=3.22\times 10^{-9} \ C$

$AP=(0.150+0.250)$

      $=0.40 \ m$

$BP=0.25 \ m$

Now,

At point P, the electric potential will be:

⇒ $V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}$

By putting values, we get

⇒     $=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]$

⇒     $=259.695 \ Volt$