Answer to Question #165787 in Physics for Susan Harriet

Question

1) Two charges, both +q, are placed 4 m apart

a) Calculate the potential due to the two charges at the point midway between them, in symbol form

(b) Calculate the potential due to the two charges at a distance of 1 m from one of the charges on the same line

c) If the electric potential was measured as 50 V at the midpoint, what will it be 1 m from one of the charges?

Answer 1

(a) The potential due to the two charges at the point midway between them is the sum of potentials due to each charge:

$$V_{midpoint}=V_1+V_2,$$ $$V_{midpoint}=\dfrac{kq}{r}+\dfrac{kq}{r}=\dfrac{2kq}{r}=\dfrac{2kq}{2}=kq=\dfrac{q}{4\pi\epsilon_0}.$$

(b)

$$V=V_1+V_2,$$ $$V=\dfrac{kq}{(\dfrac{r}{2}+1)}+\dfrac{kq}{(\dfrac{r}{2}-1)}=kq(\dfrac{1}{(\dfrac{r}{2}+1)}+\dfrac{1}{(\dfrac{r}{2}-1)}),$$ $$V=kq(\dfrac{1}{(\dfrac{4}{2}+1)}+\dfrac{1}{(\dfrac{4}{2}-1)}),$$ $$V=kq(\dfrac{1}{3}+1)=\dfrac{4}{3}kq=\dfrac{4q}{12\pi\epsilon_0}=\dfrac{q}{3\pi\epsilon_0}.$$

(c) Let's first find the charge from the formula we get in part (a):

$$q=\dfrac{V}{k}=\dfrac{50\ V}{9\cdot10^9\ \dfrac{Nm^2}{C^2}}=5.5\cdot10^{-9}\ C.$$

Then, we can find the electric potential 1 m from one of the charges from the formula we get in part (b):

$$V=\dfrac{5.5\cdot10^{-9}\ C}{3\pi\cdot8.85\cdot10^{-12}\ \dfrac{F}{m}}=66\ V.$$