Question

The drawing shows a square, each side of which has a length of L = 0.250 m. Two different positive charges q1 and q2 are fixed at the corners of the square. Find the electric potential energy of a third charge q3 = -4.00 x 10-9 C placed at corner A and then at corner B.

Answer 1

Answer:

part (a) P.E. (at A) = $-36\times \left (\dfrac{q_1}{0.25}\ +\ \dfrac{q_2}{0.35}\ \right )$

part (b) P.E. (at B) = $-36\times \left (\dfrac{q_1}{0.35}\ +\ \dfrac{q_2}{0.25}\ \right )$

Explanation:

Given,

• First charge = $q_1$
• second charge = $q_2$
• Third charge = $q_3\ =\ -4.0\times 10^{-9}\ C$
• Side of the square = a = 0.25 m

Net electric field due to the the both charges on the third charge at point A.

$P.E.\ =\ \dfrac{kq_1q_3}{a}\ +\ \dfrac{kq_2q_3}{\sqrt{2} a}\\\Rightarrow P.E.\ =\ kq_3\left (\dfrac{q_1}{a}\ +\ \dfrac{q_2}{\sqrt{2}a}\ \right )\\\Rightarrow P.E.\ =\ 9\times 10^9\times (-4.0)\times 10^{-9}\left (\dfrac{q_1}{0.25}\ +\ \dfrac{q_2}{\sqrt{2}\times 0.25}\ \right )\\\Rightarrow P.E.\ =\ -36\times \left (\dfrac{q_1}{0.25}\ +\ \dfrac{q_2}{0.35}\ \right )$

part (b)

Net electric field due to the the both charges on the third charge at point B.

$P.E.\ =\ \dfrac{kq_1q_3}{a}\ +\ \dfrac{kq_2q_3}{\sqrt{2} a}\\\Rightarrow P.E.\ =\ kq_3\left (\dfrac{q_2}{a}\ +\ \dfrac{q_1}{\sqrt{2}a}\ \right )\\\Rightarrow P.E.\ =\ 9\times 10^9\times (-4.0)\times 10^{-9}\left (\dfrac{q_2}{0.25}\ +\ \dfrac{q_1}{\sqrt{2}\times 0.25}\ \right )\\\Rightarrow P.E.\ =\ -36\times \left (\dfrac{q_2}{0.25}\ +\ \dfrac{q_1}{0.35}\ \right )$