Question

The drawing shows six point charges arranged in a rectangle. The value of q is 7.35 μC, and the distance d is 0.484 m. Find the total electric potential at location P, which is at the center of the rectangle.

Answer 1

You didn't provide the figure but I can provide you with the method to solve or the final equation so that you can put in the values and can get the final numerical result.

Now we know that Electric Potential is given by

$V= \frac{q}{4 \pi \varepsilon _o r}$

where q= charge

r= distance of the charge from a point at which Potential is to calculate.

Also since Potential is a scalar quantity thus we get added normally.

Hence Net Potential at the point P will be

  

or $V_p= \frac{q_1}{4 \pi \epsilon _o r_{1p}}+\frac{q_2}{4 \pi \epsilon _o r_{2p}}+\frac{q_3}{4 \pi \epsilon _o r_{3p}}+\frac{q_4}{4 \pi \epsilon _o r_{4p}}+\frac{q_5}{4 \pi \epsilon _o r_{5p}}+\frac{q_6}{4 \pi \epsilon _o r_{6p}}$

$V_p= \frac{1}{4 \pi \epsilon _o}[\frac{q_1}{r_{1p}}+\frac{q_2}{r_{2p}}+\frac{q_3}{r_{3p}}+\frac{q_4}{ r_{4p}}+\frac{q_5}{r_{5p}}+\frac{q_6}{r_{6p}}]$

$V_p= \frac{1}{4 \times 3.14 \times 8.85 \times 10^{-12}}[\frac{q_1}{r_{1p}}+\frac{q_2}{r_{2p}}+\frac{q_3}{r_{3p}}+\frac{q_4}{ r_{4p}}+\frac{q_5}{r_{5p}}+\frac{q_6}{r_{6p}}]$

$V_p= (8.99 \times 10^{9})[\frac{q_1}{r_{1p}}+\frac{q_2}{r_{2p}}+\frac{q_3}{r_{3p}}+\frac{q_4}{ r_{4p}}+\frac{q_5}{r_{5p}}+\frac{q_6}{r_{6p}}]$ ......................(1)

where r_1p, r_2p, r_3p, r_4p, r_5p and r_6p are the distance of all six charges respectively from point P.

Thus from the figure you can find out the distances and charges' values and can get the final numerical value by putting all in equation 1.