Question

A charge of 1mC is located at (−2,−2,2)m and a charge of 2mC is located at (2,2,2)m. A perfectly conducting medium is exists for z≤0 m and the region for z>0 is filled with air. What is the electric field at the point (2,2,4)m.

Answer 1

To calculate the electric field at the point (2, 2, 4)m due to two charges, we need to calculate the electric field due to each charge individually and then add them vectorially using the principle of superposition.

Step-by-step explanation:

The electric field at a point due to two charges can be calculated using the principle of superposition. First, we need to calculate the electric field due to each charge individually. The electric field due to a point charge is given by:

E = k * (Q / r^2)

Where E is the electric field, k is the Coulomb's constant (9.0 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the point charge. To calculate the electric field at the point (2, 2, 4)m, we can calculate the electric field due to each charge and then add them vectorially.

Electric field due to the charge of 1mC:

E1 = k * (1 * 10^-3 C) / ((2-2)^2 + (2-2)^2 + (4-2)^2)^(3/2) N/C

Electric field due to the charge of 2mC:

E2 = k * (2 * 10^-3 C) / ((2-2)^2 + (2-2)^2 + (4-2)^2)^(3/2) N/C

The net electric field is given by:

E_net = E1 + E2

By calculating E_net, we can determine the electric field at the point (2, 2, 4)m.