Question

Determine the magnitude and direction of the electric force on an electron in a uniform electric field of strength 2460 N/C2460 \mathrm{~N} / \mathrm{C} that points due east.

Solution

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Let us remember, that the electric field strength is, by definition, the force that would be applied in a unit charge; that is,

E=Fq.\vec{E}=\frac{\vec{F}}{q}.

Thus, the force on a known charge when the field is given, can be found as

F=qE.\vec{F}=q\vec{E}.

The magnitude of our force will be

F=1.610192460=3.941016.F=-1.6\cdot 10^{-19}\cdot 2460=\boxed{3.94\cdot 10^{-16}}.

The direction will be opposite that of the field, since the charge is negative. That is, the force is pointing due west.

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