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Consider the charges $q, q$, and $-q$ placed at the vertices of an equilateral triangle, as shown in Figure. What is the force on each charge?
Solution
The forces acting on charge $q$ at $A$ due to charges $q$ at $B$ and $-q$ at $C$ are $F_{12}$ along $B A$ and $F_{13}$ along $AC$ respectively, as shown in Figure.
By the parallelogram law, the total force $F _{1}$ on the charge $q$ at $A$ is given by
$F _{1}=F \hat{ r }_{1}$
where $\hat{ r }_{1}$ is a unit vector along $BC$.
The force of attraction or repulsion for each pair of charges has the same magnitude
$F=\frac{q^{2}}{4 \pi \varepsilon_{0} l^{2}}$
The total force $F _{2}$ on charge $q$ at $B$ is thus
$F _{2}=F$ $\hat{ r }_{2},$
where $\hat{ r }_{2}$ is a unit vector along $AC.$
Similarly the total force on charge $-q$ at $C$ is $F _{3}=\sqrt{3} F$ in , where $\hat{ n }$ is the unit vector along the direction bisecting the $\angle BCA$.
It is interesting to see that the sum of the forces on the three charges is zero, i.e., $F _{1}+ F _{2}+ F _{3}= 0$
The result is not at all surprising. It follows straight from the fact that Coulomb's law is consistent with Newton's third law.
