Three point charges of magnitude 5 μ C , 0 . 16 μ C and 0 . 3 μ C are located at the vertices A , B , C of a right angled triangle whose sides are A B = 3 cm , B C = 3 2 cm and C A = 3 cm and point A

Question

Three point charges of magnitude 5 μ C , 0 . 16 μ C and 0 . 3 μ C are located at the vertices A , B , C of a right angled triangle whose sides are A B = 3 cm , B C = 3 2 cm and C A = 3 cm and point A is the right angle corner. Charge at point A experiences _____ N of electrostatic force due to the other two charges.

TestHub · Accepted Answer

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

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