Question 26 of 311

Three point charges of magnitude $5\mu C,0.16\mu C$ and $0.3\mu C$ are located at the vertices $A,B,C$ of a right angled triangle whose sides are $AB=3cm$, $BC=3\sqrt{2}cm$ and $CA=3cm$ and point $A$ is the right angle corner. Charge at point $A$ experiences _____ $N$ of electrostatic force due to the other two charges.

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Answer: 17

Using Coulomb's law the electrostatic force on $A$ due to $C$
${F}_{AC}=\dfrac{K{q}_{A}{q}_{C}}{{\left({r}_{AC}\right)}^{2}}=\dfrac{k\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$$=\dfrac{9\times {10}^{9}\times 5\times 0.3\times {10}^{-12}}{9\times {10}^{-4}}$
$=1.5\times 10=15N$
Similarly, the force on $A$ due to $B$
${F}_{AB}=\dfrac{K{q}_{A}{q}_{B}}{{\left({r}_{AB}\right)}^{2}}=\dfrac{9\times {10}^{9}\times 5\times 0.16\times {10}^{-12}}{9\times {10}^{-4}}=8N$
We can see that the forces are perpendicular to each other. Therefore, the net force experienced by charge at $A$ is,
$F=\sqrt{{{F}_{AC}}^{2}+{{F}_{AB}}^{2}}$$=\sqrt{{15}^{2}+{8}^{2}}=\sqrt{289}=17N$