Answer to Question #87948 in Electricity and Magnetism for Nicholas Legault

Question

What electric potential would be measured at a distance of 12cm from a charge of 3.5 x 10-5 C? What would be the potential if the distance were increased to 30 cm?
How much work must be done to carry a 3.0 x 10-5 C charge from 12 cm to 30 cm?

Answer 1

A) By the definition of the electric potential, we have:

$$V = k \dfrac{q}{r},$$

here, $k = 9 \cdot 10^9 \dfrac{N \cdot m^2}{C^2}$ is the Coulomb's constant, $q =3.5 \cdot 10^{-5} C$ is the charge, $r$ is the distance from the charge to the point where the potential is measured.

Then, we get:

$$V_1 = k \dfrac{q}{r_1} = 9 \cdot 10^9 \dfrac{N \cdot m^2}{C^2} \cdot \dfrac{3.5 \cdot 10^{-5} C}{0.12 m} = 2.625 \cdot 10^6 V.$$ $$V_2 = k \dfrac{q}{r_2} = 9 \cdot 10^9 \dfrac{N \cdot m^2}{C^2} \cdot \dfrac{3.5 \cdot 10^{-5} C}{0.30 m} = 1.05 \cdot 10^6 V.$$

B) By the definition, the change in the potential energy of a charged particle is equal to the negative of the work done by the external force:

$$\Delta PE = PE_2 - PE_1 = -W,$$ $$W = qV_2 - qV_1,$$ $$W = 3.5 \cdot 10^{-5} C \cdot (1.05 \cdot 10^6 V - 2.625 \cdot 10^6 V) = -55.125 J.$$

Answer:

A) $V_1 = 2.625 \cdot 10^6 V, V_2 = 1.05 \cdot 10^6 V.$

B) $W = 55.125 J.$