Question

3 equal charges which has magnitude of 2×10^6 coulomb is placed at the corners of right angled triangle of sides 3, 4, 5

find the force on the charge at right angle corner.

Answer 1

See the diagram. 
the force acting on charge at B due to charge at A is
$F_1= \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r^2} =9*10^9* \frac{(2*10^6)^2}{4^2}=2.25*10^{21}N$
It acts in the direction from A to B.

the force acting on charge at B due to charge at C is
$F_2= \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r^2} =9*10^9* \frac{(2*10^6)^2}{3^2}=4*10^{21}N$
It acts in the direction from C to B.

Since F₁ and F₂ are at right angles to each other, the resultant F is given by

$F= \sqrt{F_1^2+F_2^2}\\F = \sqrt{(2.25*10^{21})^2+(4*10^{21})^2} \\F= \sqrt{21.0625*10^{42}} \\F=\boxed{4.59*10^{21}\ N}$