Physics 2007 JAMB Past Questions - Question 23: A charge 50 µC has an electric field strength of 360 NC-¹ at a certain point. The electric field strength due to another charge 120 µC kept at the same distance apart and in the same medium is.

Correct Answer

C

Explanation

To find the electric field strength due to another charge 120 µC at the same distance apart and in the same medium, you can use the principle of superposition. The electric field from multiple charges at a point adds vectorially. The formula to calculate the electric field strength at a point due to a point charge is:

E = k * (|q| / r^2)

Where:
- E is the electric field strength at the point.
- k is the electrostatic constant (approximately 8.99 x 10^9 Nm^2/C^2).
- |q| is the magnitude of the charge.
- r is the distance from the charge to the point where you want to find the electric field strength.

In this case, you have two point charges:

1. The first charge is 50 µC = 50 x 10^(-6) C.
2. The second charge is 120 µC = 120 x 10^(-6) C.

You mentioned that the electric field strength due to the first charge is 360 NC^(-1) (360 N/C).

To find the distance (r) from the first charge to the point where you want to find the electric field, you need more information about the geometry of the setup. If the distance is the same for both charges, you can use the electric field formula to find the distance (r). Then, you can use this distance to find the electric field strength due to the second charge:

E2 = k * (|q2| / r^2)

Substituting the values:
E2 = (8.99 x 10^9 Nm^2/C^2) * (120 x 10^(-6) C) / (r^2)

Once you find the value of r, you can calculate E2. If the distance is different for the second charge, you would need to know that distance as well to calculate E2 accurately.