(A B C ) is a right angled triangle in which (A B=3 ~cm , B C=4 ~cm ) and right angle is at (B ). Three charges (+15 C ,+12 C ) and (-20 C ) are placed respectively at (A, B ) and (C ). The force acting on the charge at (B ) is

Step-by-step solution

According to the question, 3 charge particles are placed at the vertices of a right angle triangle (A B C ) as shown in the figure below, Where, ( aligned & Q_A=+15 C , & Q_B=12 C & Q_C=-20 C aligned ) and (Q_C=-20 C ) Now, the force, (F_ A B = k Q_A Q_B r_ A B ^2 ) ( F_ A B = k 15 12 10⁻¹² 9 10⁻⁴ =k 20 10⁻⁸ ~N ) Similarly, (F_ B C = k 20 12 10⁻¹² 16 10⁻⁴ =k 15 10⁻⁸ ~N ) Now, the resultant, (F_B= F_ A B ^2+F_ B C ^2 ( =90^ ) ) (F_B=9 10 [ 20^2+15^2 ]=2250 ~N ) Hence, the force acting on the charge at point (B ) is