Question

What is the electric potential at a point midway between two charges, -5.37 microC and -4.11 microC, separated by 8.43 cm?

Answer 1

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.

Step-by-step explanation:

The potential V is given by:

`V=\frac{{q}}{{{4\pi\epsilon_0d}}}`

The charge q1 is at a distance of 4.215 cm from the midpoint.

The potential due to charge q1 is:

`V_1=(q1)/(4\pi\epsilon_0d)=\frac{-5.37*10^(-6)\text{ C}*9*10^9\text{ Nm}^2\text{/C}^2}{4.215*10^(-2)\text{ m}}=-1146619.21\text{ V}=-1.146*10^6\text{ V}`

The charge q2 is at a distance of 4.215 cm from the midpoint.

The potential due to the charge q2 is:

`V_2=(q_2)/(4\pi\epsilon_0d)=\frac{-4.11*10^(-6)\text{ C}*9*10^9\text{ Nm}^2\text{/C}^2}{4.215*10^(-2)\text{ m}}=-877580.07\text{ V}=-0.877*10^6\text{ V}`

The electric potential V at a midpoint is:

`V=V_1+V_2=-1.146*10^6\text{ V}-0.877*10^6\text{ V}=-2.023*10^6\text{ V}=-2.023\text{ MV}`

Final answer:

The electric potential at a point midway between the given two charges is -2.023 MV.