Answer to Question #126457 in Physics for keerthana

Question

Find the electric potential at the midpoint of two point charges +8μC and -16μC separated by 0.5 m in vacuum. At what point between them the potential is zero?

Answer 1

The electric potential of the point charge in some point of space is given by the following expression:

$$\phi = k\dfrac{q}{r}$$

where $q$ is the charge, $r$ is the distance fpom the point to the charge and $k = 9\times 10^{9} N\cdot m^2/C^2$ is the Coulomb's constant.

Also note the fact, that the resul potential from the two charges is the sum of their individual potentials. Thus, for the midpoint obtain:

$$\phi =\phi_1 + \phi_2 = k\dfrac{q_1}{r} + k\dfrac{q_2}{r} = \dfrac{k}{r}(q_1 +q_2)$$

where $r = 0.5m/2 = 0.25m$.

Thus, obtain:

$$\phi= \dfrac{k}{r}(q_1 +q_2) = \dfrac{9\times10^9}{0.25}(8\times10^{-6} -16\times10^{-6}) = -288\times10^3V$$

Let's find the point $r_0$ (counting from the first charge) of zero potential.

$$\phi= k\dfrac{q_1}{r_0} + k\dfrac{q_2}{r-r_0} = 0$$

where $r = 0.5m$.

Expressing the $r_0$, obtain:

$$r_0 = \dfrac{q_1r}{q_1-q_2} = \dfrac{8\times10^{-6}\cdot 0.5}{8\times10^{-6}-(-16)\times10^{-6}} \approx 0.17m$$

Answer. (a) -288 kV, (b) 0.17 m counting from the first charge.