Answer:48 V
Explanation:
Given
Three charged particle with charge
[tex]q_1=50\ nC at y=6\ m[/tex]
[tex]q_2=-80\ nC at x=-4\ m[/tex]
[tex]q_3=70\ nC at y=-6\ m[/tex]
Electric Potential is given by
[tex]V=\frac{kQ}{r}[/tex]
Distance of [tex]q_1[/tex] from [tex]x=8\ m[/tex]
[tex]d_1=\sqrt{6^2+8^2}[/tex]
[tex]d_1=\sqrt{36+64}[/tex]
[tex]d_1=10\ m[/tex]
similarly [tex]d_2=8-(-4)[/tex]
[tex]d_2=12\ m[/tex]
[tex]d_3=\sqrt{(-6)^2+8^2}[/tex]
[tex]d_3=\sqrt{36+64}[/tex]
[tex]d_3=10\ m[/tex]
Potential at [tex]x=8\ m[/tex] is
[tex]V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}[/tex]
[tex]V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}][/tex]
[tex]V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}[/tex]
[tex]V_{net}=9\times 5.33[/tex]
[tex]V_{net}=47.97\approx 48\ V[/tex]
