Answer:
71.06 × 10⁻³ N.m
0
Explanation:
Given that:
four point charges of magnitude = 2.7 mu
at the corners of a square of side 5m
a) Find the electrostatic potential energy if all of the charges are negative.
Electrostatic potential energy can be calculated using the formula:
P.E = [tex]E(\frac{kq_1q_2}{r})[/tex]
If all the charges are negative; then, we have:
P.E = [tex][\frac{q_1q_2}{5m}]+[\frac{kq_1q_2}{5m}]+[\frac{kq_1q_3}{5\sqrt{2m} }]+[\frac{kq_2q_4}{5\sqrt{2m} } ]+{\frac{kq_2q_3}{5m}+[\frac{kq_3q_4}{5m}][/tex]
P.E = [tex][\frac{k(-q)(-q)}{5m}] +[\frac{k(-q)(-q)}{5m}] +[\frac{k(-q)(-q)}{5m\sqrt{2}}] +[\frac{k(-q)(-q)}{5m\sqrt{2}}] +[\frac{k(-q)(-q)}{5m}] +[\frac{k(-q)(-q)}{5m}][/tex]
P.E = [tex]kq^2[\frac{1}{5}+ \frac{1}{5}+\frac{1}{5\sqrt{2} }+\frac{1}{5\sqrt{2} }+\frac{1}{5}+\frac{1}{5}][/tex]
P.E = [tex]kq^2[\frac{4}{5}+\frac{2}{5\sqrt{2} } ]m^{-1[/tex]
P.E = kq² (1.083)m⁻¹
P.E = (9 × 10⁹ Nm²/C²)(2.7 × 10⁻⁶ C)²(1.083 m⁻¹)
P.E = 0.07106 N.m
P.E = 71.06 × 10⁻³ N.m
b) Find the electrostatic potential energy if three of the charges are positive and one is negative.
P.E = [tex]E(\frac{kq_1q_2}{r})[/tex]
If three of the charges are positive and one is negative; we have
P.E = [tex][\frac{q_1q_2}{5m}]+[\frac{kq_1q_2}{5m}]+[\frac{kq_1q_3}{5\sqrt{2m} }]+[\frac{kq_2q_4}{5\sqrt{2m} } ]+{\frac{kq_2q_3}{5m}+[\frac{kq_3q_4}{5m}][/tex]
P.E = [tex][\frac{k(q)(q)}{5m}] +[\frac{k(q_1)(-q)}{5m}]+[\frac{k(q)(q)}{5m\sqrt{2} }]+[\frac{k(q_2)(-q)}{5m\sqrt{2} }]+[\frac{k(q)(q)}{5m}]+[\frac{k(q)(-q)}{5m}][/tex]
P.E = [tex]kq^2[\frac{1}{5}+ \frac{1}{5}+\frac{1}{5\sqrt{2} }-\frac{1}{5\sqrt{2} }+\frac{1}{5}+\frac{1}{5}]m^{-1[/tex]
P.E = [tex]kq^2[0][/tex]
P.E = 0
