Calculating Electric Force: 0.15 N in +y Direction

[samsungman]

I need help with this problem.

What is the magnitude and direction of the elctric force on a -1.2 micro Coulomb charge at a point where the electric field is 2500 N/C and is directed along the +y axis?

It has something to do with this equation:
$$F=k_e \frac {Q_1 Q_2} {r^2}$$

I got 0.15 N, in the +y direction

 

[jamesrc]

That equation will not help you. There is an equation relating the force felt by a charge in an electric field to the magnitude of the charge and the magnitude of the electric field. In your class, it may have been called the Lorentz force law. Does that help?

 

[m2003]

as well.

Based on the given information and the equation provided, we can calculate the electric force on the -1.2 micro Coulomb charge using the formula F = k_e * (Q1*Q2)/r^2, where k_e is the Coulomb's constant, Q1 is the charge of the -1.2 micro Coulomb charge, Q2 is the charge of the point where the electric field is acting, and r is the distance between the two charges.

Substituting the values, we get F = (9*10^9 N*m^2/C^2) * (-1.2*10^-6 C * 0 C)/(r^2) = -10.8 N/r^2.

Since we are only interested in the magnitude and direction of the electric force, we can ignore the negative sign and focus on the magnitude. The magnitude of the electric force is 10.8 N divided by the distance squared between the two charges.

Since the electric field is directed along the +y axis, we can assume that the distance between the two charges is in the +y direction as well. Therefore, the direction of the electric force will also be in the +y direction.

In conclusion, the magnitude of the electric force on the -1.2 micro Coulomb charge is 10.8 N in the +y direction.