A man is standing beneath two rain clouds, equidistant from each cloud, as shown in the figure. Cloud A has an excess of 7.1 104 negative charges while cloud B has 7.1 104 positive charges (where a charge has magnitude e).
(a) What is the magnitude of the net electric field where the man is standing?
N/C
(b) What is the direction of the net electric field where the man is standing? (Assume the +x-axis is to the right.)
° counterclockwise from the +x-axis
To find the magnitude and direction of the net electric field where the man is standing, we can use the principle of superposition. The principle states that the net electric field at a certain point is the vector sum of the electric fields due to individual charges or charge distributions.
In this case, we have two clouds with excess charges. Cloud A has negative charges and Cloud B has positive charges. Let's break down the steps to find the answers:
(a) Magnitude of the net electric field:
1. Determine the electric field due to each cloud separately. The electric field due to a point charge is given by the equation:
E = k * (q / r^2)
where E represents the electric field, k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2), q is the magnitude of the charge, and r is the distance from the charge.
2. Since the man is equidistant from both clouds, the magnitudes of the electric fields due to each cloud will be the same. Let's call this magnitude E_cloud.
E_cloudA = k * (7.1 x 10^4 e / r^2)
E_cloudB = k * (7.1 x 10^4 e / r^2)
Note that the excess charge magnitude (7.1 x 10^4) is given in terms of the elementary charge (e).
3. The net electric field at the man's position is the vector sum of the electric fields due to each cloud:
E_net = E_cloudA + E_cloudB
Since the magnitudes of the electric fields are the same, we can simplify this to:
E_net = 2 * E_cloudA
4. Substitute the appropriate values into the equation and calculate E_net.
(b) Direction of the net electric field:
1. The direction of the net electric field can be determined using the principle of superposition. The direction will be counterclockwise from the +x-axis.
2. Since the man is standing equidistant from the clouds, the individual electric fields will be of equal magnitude and opposite in direction.
3. By adding these two oppositely directed vectors, the resultant vector (E_net) will point in the counterclockwise direction from the +x-axis.
Following these steps will help you find the magnitude and direction of the net electric field at the man's position. Don't forget to substitute the given values and double-check your calculations.