Respuesta :
Given:
q1 = -5.37 microC
q2 = -4.11 microC
d = 8.43 cm
To find:
The electric potential at a point midway between two charges.
Explanation:
The potential V is given by:
[tex]V=\frac{{q}}{{{4\pi\epsilon_0d}}}[/tex]The charge q1 is at a distance of 4.215 cm from the midpoint.
The potential due to charge q1 is:
[tex]V_1=\frac{q1}{4\pi\epsilon_0d}=\frac{-5.37\times10^{-6}\text{ C}\times9\times10^9\text{ Nm}^2\text{/C}^2}{4.215\times10^{-2}\text{ m}}=-1146619.21\text{ V}=-1.146\times10^6\text{ V}[/tex]The charge q2 is at a distance of 4.215 cm from the midpoint.
The potential due to the charge q2 is:
[tex]V_2=\frac{q_2}{4\pi\epsilon_0d}=\frac{-4.11\times10^{-6}\text{ C}\times9\times10^9\text{ Nm}^2\text{/C}^2}{4.215\times10^{-2}\text{ m}}=-877580.07\text{ V}=-0.877\times10^6\text{ V}[/tex]The electric potential V at a midpoint is:
[tex]V=V_1+V_2=-1.146\times10^6\text{ V}-0.877\times10^6\text{ V}=-2.023\times10^6\text{ V}=-2.023\text{ MV}[/tex]Final answer:
The electric potential at a point midway between the given two charges is -2.023 MV.
