Question

Three charges A(4 µC), B(-6 µC) and C(2 µC) are placed at the vertices of a right angle triangle ABC. AC=10 cm, BC=6 cm. Find net force on charge B due to C and A charges

Answer 1

For the arrangement of charges, by Coulomb's law, we have:

$$F_{AC}=\frac{kq_Aq_C}{r_{AC}^2}=7.2\text{ N},\\\space\\ F_{BC}=\frac{kq_Bq_C}{r_{BC}^2}=30\text{ N},\\\space\\ F_\text{net}=\sqrt{F_{AC}^2+F_{AC}^2}=30.8\text{ N}.$$