Question
Three equal charges, $2.0 \times 10^{-6}C$ each, are held fixed at the three corners of an equilateral triangle of side 5cm. Find the Coulomb force experienced by one of the charges due to the rest two.
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Answer
Three charges are held at three corners of a equilateral trangle.
Let the charges be A, B and C.
It is of length 5cm or 0.05m
Force exerted by B on A $= F_1$
Force exerted by C on A $= F_2$
So, force exerted on A = resultant $F_1 = F_2$
$\Rightarrow\text{F}=\frac{\text{kq}_2}{\text{r}^2}$
$=\frac{9\times10^9\times2\times2\times2\times10^{-12}}{5\times5\times10^{-4}}$
$=\frac{36}{25}\times10$
$=14.4$
Now, force on $A = 2 × F$ cos $30^\circ$ since it is equilateral $\triangle.$
$⇒ $Force on $\text{A}=2\times1.44\times\sqrt{\frac{3}{2}}$
$=24.94\text{N}.$
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