A uniform electric field is directed upward and has a magnitude of 5 N/C. What are the magnitude and direction of the force on a charge of -3 C placed in this field?
F=Eq
E=5
q=-3
5(-3)=-15N Downward
Ok I have the downward correct the -15 is incorrect. Did I use the wrong equation?
Your F = E q formula is correct. Are you sure the charge was 3 Coulombs and not 3 microcoulombs (3 uC) ?
I'm sure... I copy/paste this question on here to be sure I got it right.
Yes, you used the correct equation for calculating the force on a charge placed in an electric field. However, the magnitude of the force you calculated is incorrect. Let's re-calculate it correctly:
Given:
Electric field magnitude (E) = 5 N/C
Charge (q) = -3 C
The formula to calculate the force (F) on a charge in an electric field is F = Eq.
Substituting the given values:
F = (5 N/C) * (-3 C)
F = -15 N
So, the correct magnitude of the force is 15 N. However, the direction of the force will be opposite to the direction of the field since the charge is negative. Therefore, the direction of the force will be downward.
To find the magnitude and direction of the force on a charged particle in a uniform electric field, you can use the equation F = qE, where F is the force on the particle, q is the charge of the particle, and E is the electric field strength.
In this case, you correctly identified the electric field strength (E) as 5 N/C and the charge (q) as -3 C. Therefore, using the formula F = qE, you should multiply the charge and the electric field strength to get the force:
F = (-3 C)(5 N/C) = -15 N
You are correct that the force is downward because the charge is negative. However, it's important to note that the magnitude of the force (15 N) is always positive, regardless of the direction. Therefore, the magnitude of the force on the charge is 15 N, and the direction of the force is downward.