Answer to Question #250486 in Physics for Soriano

Question

Particles A and B carrying charge 3.0nC are at the base corners of an equilateral triangle 2.8 m on a side

What is the potential[relative to zero at infinity] at the apex of the triangle if particle B is replaced by a particle carrying a charge of -3.0nC?

Answer 1

Since the triangle is equilateral, the electrostatic potential produced at the apex by the particle A is given as follows:

$$\phi_A = k\dfrac{q_A}{r}$$

where $k$ is Coulomb constant, $q_A$ is the charge of A, and $r$ is the side of the triangle.

Similarly, the electrostatic potential produced at the apex by the particle B is given as follows:

$$\phi_B = k\dfrac{q_B}{r}$$

But, since in the later case $q_A = -q_B$, the total potential will be:

$$\phi = \phi_A + \phi_B = k\dfrac{q_A}{r} + k\dfrac{q_B}{r} = k\dfrac{q_A}{r} - k\dfrac{q_A}{r} = 0$$

Answer. 0.