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Three equal charges, each having a magni...

Three equal charges, each having a magnitude of `2 xx 10^(-6)"C"`, are placed at the three corners of a right-angled triangle of sides 3, 4, and 5 cm. Find the force on the charge at the right-angled corner.

Text Solution

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The force on A due to B is
`F_1 = (1)/( 4pi epsilon_(0) ) ((q_A xx q_B)/( d^2) )`
`F_1 = (9 xx 10^9 xx 2 xx 10^(-6) xx 2 xx 10^(-6) )/( (4 xx 10^(-2) )^2)`
`= (9 xx 4 xx 10^(-3) )/( 16 xx 10^(-4) ) = 22.5`N
This force acts along BA. Similarly, the force on A due to C is `F_2 = 40` N in the direction of CA. Thus, the net electric force on A is
`F = sqrt(F_1^2 + F_2^2) = sqrt((22.5)^2 + (40)^2)`
`=sqrt( 506.25 + 1600)`
`F = 45.9 N.`
This ,resultant makes an angle `theta ` with BA, where `tan theta = (40)/(22.5) = (16)/(9)`.
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