A charge of -4.50 10-6 C is placed at a point in space where the electric field is directed toward the right and has a magnitude of 9.00 104 N/C. What are the magnitude and direction of the electrostatic force on this charge?
force=Eq
look at direction first, E is to the right (E is the direction a + test charge will do, you are adding a neg charge), so the force will be to the left
To find the magnitude of the electrostatic force, we can use the formula:
F = q * E
where,
F is the electrostatic force,
q is the charge, and
E is the electric field.
Given:
q = -4.50 x 10^(-6) C (charge)
E = 9.00 x 10^4 N/C (electric field)
Plugging in the given values:
F = (-4.50 x 10^(-6) C) * (9.00 x 10^4 N/C)
Calculating the multiplication:
F = -4.05 x 10^(-1) N/C
Therefore, the magnitude of the electrostatic force on the charge is 4.05 x 10^(-1) N.
To determine the direction of the electrostatic force, we need to consider the sign of the charge. Since the charge is negative (-4.50 x 10^(-6) C), the force will be in the opposite direction to the electric field.
Therefore, the direction of the electrostatic force is toward the left.
To find the magnitude and direction of the electrostatic force on a charge, we can use the equation:
F = q * E
Where F is the force, q is the charge, and E is the electric field.
In this case, the charge q is given as -4.50 x 10^-6 C and the electric field E is given as 9.00 x 10^4 N/C.
First, let's calculate the magnitude of the force:
F = (-4.50 x 10^-6 C) * (9.00 x 10^4 N/C)
= -4.05 x 10^-1 N
The magnitude of the force is 4.05 x 10^-1 N.
Next, let's determine the direction of the force. Since the electric field is directed toward the right, and the charge is negative, the force will be in the opposite direction of the electric field.
Therefore, the direction of the force is toward the left.
So, the magnitude of the force is 4.05 x 10^-1 N, and the direction is toward the left.