A charge of +3.7 ✕ 10-5 C is located at a place where there is an electric field that points due east and has a magnitude of 16400 N/C. What are the magnitude and direction of the force acting on the charge?
To find the magnitude and direction of the force acting on the charge, we can use the formula for the force experienced by a charged particle in an electric field:
F = q * E
where F is the force, q is the charge, and E is the electric field.
In this case, the given charge is +3.7 ✕ 10^(-5) C, and the electric field has a magnitude of 16400 N/C and points due east.
Substituting the given values into the formula, we have:
F = (3.7 ✕ 10^(-5) C) * (16400 N/C)
F = 6.068 ✕ 10^(-1) N
Therefore, the magnitude of the force acting on the charge is approximately 0.607 N.
To determine the direction of the force, we need to consider the sign of the charge. Since the charge is positive, it will experience a force in the same direction as the electric field. Thus, the force is directed due east.
So, the magnitude of the force is 0.607 N, and its direction is due east.