What is the relationship between the direction of electric field and that of the electric force on a test charge group of answer choices?

The given equation $\vec E=\frac{\vec F}q$ can be easily rearranged to $\vec F=\vec Eq$.

When the charge $q$ is positive then the direction of $\vec F$ is the same as the direction of $\vec E$ - all values are positive.

When the charge $q$ is negative (i.e. $-q$) then $\vec F$ will be negative: $\vec F=\vec Eq \Rightarrow -\vec F=\vec E(-q)$.

This is similar to the attraction of charges $F=\frac{kq_1q_2}{r^2}$ where, if $q_1$ and $q_2$ both have the same sign (both positive or both negative) then $F$ is positive and the charges repel each other, whereas if $q_1$ and $q_2$ have opposite signs (one positive and one negative) then $F$ is negative and the charges are attracted to each other.

But instead of two charges, you have one charge and an electric field. If you consider the electric field to be originating from a positive charge ($+q$) then the direction of the field is away from that (imaginary) charge and the direction of force exerted on another positive charge in the field is in the same direction as the field - the source of the field repels the positive charge, but the direction of force exerted on a negative charge in the field is instead in the opposite direction to the field - the source of the field attracts the negative charge.

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