12.1.2 Electric Force between Two Point Charges

Coulomb’s Law states that the electrical force F_e 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

Q_A, Q_B and Q_C 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 Q_C.

Solution

The electric force exerted by Q_A on Q_C,

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

The electric force exerted by Q_B on Q_C,

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

Note that Q_C is attracted towards Q_A, but repelled away from Q_B. F_A and F_B are thus in the directions shown in the diagram. The resultant electric force acting of Q_C is thus given by the vector summation of F_A and F_B. By inspection, the vector summation triangle is an equilateral triangle. So F_R turns out to be a leftward force of 4.405 mN.