Three equal charges which have a magnitude of 2 x 10-6 coulomb is placed at the corners of a right-angled triangle of sides 3, 4, 5 cm.  Find the force on the charge at right angle corner.

Answers (1)
S safeer

F_1=k\frac{q^2}{4^2}=9\times10^9\frac{(2\times10^{-6})^2}{4^2}=\frac{9}{4}\times10^{-3}N

F_2=k\frac{q^2}{r^2}=9\times10^9\frac{(2\times10^{-6})^2}{3^2}={4}\times10^{-3}N

\\F=\sqrt{F_1^2+F_2^2}=\sqrt{\frac{81}{16}+16}\times10^{-3}\\\\=4.6\times10^{-3}\times10^{4}N=46N

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