Physics 2021 JAMB Past Questions - Question 32: The potential at a point distance 0.03m from a small charge q is -1.0 × 10µV. What is the magnitude of q? [Assume 1/(4πÔ‘o ) = 9 × 10¹Nm²C-²²].

Correct Answer

A

Explanation

The potential V at a distance r from a point charge q is given by the equation V = k * (q / r), where k is the Coulomb's constant (8.99 × 10^9 N m^2/C^2).

In this case, we have the potential V = -1.0 × 10^5 V, the distance r = 0.03 m, and we need to find the magnitude of q.

Using the equation V = k * (q / r), we can rearrange it to solve for q:

q = V * r / k

Plugging in the values, we get:

q = (-1.0 × 10^5 V) * (0.03 m) / (8.99 × 10^9 N m^2/C^2)
  ≈ -0.334 × 10^-5 C

So, the magnitude of q is approximately 0.334 × 10^-5 C.