Electric Force

--Asaxon7 (talk) 00:48, 18 November 2015 (EST) Claimed by Alayna Saxon

This page contains information on the electric force on a point charge. Electric force is created by an external Electric Field.

The Coulomb Force Law

The formula for the magnitude of the electric force between two point charges is:

[math]\displaystyle{ |\vec F|=\frac{1}{4 \pi \epsilon_0 } \frac{|{q}_{1}{q}_{2}|}{r^2} }[/math]

where [math]\displaystyle{ {q}_{1} }[/math] and [math]\displaystyle{ {q}_{2} }[/math] are the magnitudes of charge of point 1 and point 2 and [math]\displaystyle{ r }[/math] is the distance between the two point charges. The units for electric force are in Newtons.

Direction of Electric Force

The electric force is along a straight line between the two point charges in the observed system. If the point charges have the same sign (i.e. both are either positively or negatively charged), then the charges repel each other. If the signs of the point charges are different (i.e. one is positively charged and one is negatively charged), then the point charges are attracted to each other.

Derivations of Electric Force

The electric force on a particle can also be written as:

[math]\displaystyle{ \vec F=q\vec E }[/math]

where [math]\displaystyle{ q }[/math] is the charge of the particle and [math]\displaystyle{ \vec E }[/math] is the external electric field.

This formula can be derived from [math]\displaystyle{ |\vec F|=\frac{1}{4 \pi \epsilon_0 } \frac{|{q}_{1}{q}_{2}|}{r^2} }[/math], the electric force between two point charges. The magnitude of the electric field created by a point charge is [math]\displaystyle{ |\vec E|=\frac{1}{4 \pi \epsilon_0 } \frac{|q|}{r^2} }[/math], where [math]\displaystyle{ q }[/math] is the magnitude of the charge of the particle and [math]\displaystyle{ r }[/math] is the distance between the observation location and the point charge. Therefore, the magnitude of electric force between point charge 1 and point charge 2 can be written as:

[math]\displaystyle{ |\vec F|=\frac{1}{4 \pi \epsilon_0 } \frac{|{q}_{1}{q}_{2}|}{r^2}=|{q}_{2}|\frac{1}{4 \pi \epsilon_0 } \frac{|{q}_{1}|}{r^2}=|{q}_{2}||\vec{E}_{1}| }[/math]

The units of charge are in Coulombs and the units for electric field are in Newton/Coulombs, so this derivation is correct in its dimensions since multiplying the two units gives just Newtons. The Newton is the unit for electric force.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Problem 1

Problem: Find the magnitude of electric force on two charged particles located at [math]\displaystyle{ \lt 0, 0, 0\gt }[/math]m and [math]\displaystyle{ \lt 0, 10, 0\gt }[/math]m. The first particle has a charge of +5 C and the second particle has a charge of -10 C. Is the force attractive or repulsive?

Step 1: Find the distance between the two point charges.

[math]\displaystyle{ d=\sqrt{(0 m-0 m)^2+(0 m-10 m)^2+(0 m-0 m)^2}=\sqrt{100 m} }[/math]

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References

Matter & Interactions, Vol. II: Electric and Magnetic Interactions, 4th Edition