12.1.2 Electric Force between Two Point Charges

Coulomb’s Law states that the electrical force Fe between two stationary point charges Q1 and Q2 separated by distance d is given by the formula

\displaystyle {{F}_{e}}=k\frac{{{{Q}_{1}}{{Q}_{2}}}}{{{{d}^{2}}}}

Doesn’t it look strikingly similar to \displaystyle {{F}_{g}}=G\frac{{{{M}_{1}}{{M}_{2}}}}{{{{d}^{2}}}}? The point masses M1 M2 are replaced by point charges Q1 Q2. The gravitational constant \displaystyle G=6.67\times {{10}^{{-11}}}\text{ N }{{\text{m}}^{2}}\text{ k}{{\text{g}}^{{-2}}} is replaced by the Coulomb’s constant \displaystyle k=\frac{1}{{4\pi {{\varepsilon }_{0}}}}=8.99\times {{10}^{9}}\text{ N }{{\text{m}}^{2}}\text{ }{{\text{C}}^{{-2}}}. Just like Newton’s law of gravitation, Coulomb’s law also obeys the inverse-square law. Furthermore, it is also a manifestation of N3L. No wonder people joke that Coulomb copied Newton’s homework.

The one major difference between gravitation and electricity is that while there is only one type of mass, there are two types of charges: positive and negative. While gravitational forces are always attractive, electric forces can be attractive (between two unlike charges) or repulsive (between two like charges).

To include the sign or not?

One thing to note. The formula \displaystyle {{F}_{e}}=k\frac{{{{Q}_{1}}{{Q}_{2}}}}{{{{d}^{2}}}} is usually used to calculate the magnitude of the force only. There is no point including the signs of the charges when applying this formula. It is better and easier to figure out the direction of the forces (whether they are leftward, rightward, upward, downward etc) from the context (rather than from the formula).

Example

QA, QB and QC are point charges of +2.1 pC, −2.1 pC and −2.1 pC, bound at the vertices of an equilateral triangle of side 3.0 um as shown in the diagram below. Calculate the force experienced by QC.

Solution

The electric force exerted by QA on QC,

\displaystyle \begin{aligned}{{F}_{A}}&=k\frac{{{{Q}_{A}}{{Q}_{C}}}}{{{{d}^{2}}}}\\&=(8.99\times {{10}^{9}})\frac{{(2.1\times {{{10}}^{{-12}}})(2.1\times {{{10}}^{{-12}}})}}{{{{{(3.0\times {{{10}}^{{-6}}})}}^{2}}}}\\&=4.405\text{ mN}\end{aligned}

The electric force exerted by QB on QC,

\displaystyle \begin{aligned}{{F}_{B}}&=k\frac{{{{Q}_{B}}{{Q}_{C}}}}{{{{d}^{2}}}}\\&=(8.99\times {{10}^{9}})\frac{{(2.1\times {{{10}}^{{-12}}})(2.1\times {{{10}}^{{-12}}})}}{{{{{(3.0\times {{{10}}^{{-6}}})}}^{2}}}}\\&=4.405\text{ mN}\end{aligned}

Note that QC is attracted towards QA, but repelled away from QB. FA and FB are thus in the directions shown in the diagram. The resultant electric force acting of QC is thus given by the vector summation of FA and FB. By inspection, the vector summation triangle is an equilateral triangle. So FR turns out to be a leftward force of 4.405 mN.

Video Explanation

Coulomb’s Copycat Law

Applets

Electric Field of Dreams (PhET)

Concept Test

2201

Interesting

Electric vs Gravitation

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