Three equal charges, each having a magnitude of. ` 2.0 xx 10 ^ (-6) C`, are placed at the thre corners of a right. angled triangle of sides 3 cm, 4cm and 5 cm. the force (magnitude) on the charge at the right angled corner is

Explanation

`The situation is shown in figure. The force on
A dueto B is
` F_1 ((2.0 xx 10(-6)C))(2.0 xx 10(-6)C))/4piepsilon0(4cm)^2`
=` 9x10 ^ 9 Nm^2 C(-2) x 4 x 10 ^ (-12) C^2 X 1)/ (16xx10^(-4) m^2)`
= 22.5 N.
The 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(22.5N)^2 + (40 N) N^2 = 45.9 N`
. This resultant makes an angle 0 with BA where
`tan theta = 40/22.5 = 16.9.`