Question

Three equal charges of 10 x 10-8 C respectively each located at the corners of a right angled triangle whose sides are 15cm, 20cm and 25cm respectively. Find the force exerted on the charge located at the 90 degree angle.​

Answer 1

Given :

Charge (q) = 10 × 10⁻⁸ C

Sides of the triangle = 15cm, 20cm and 25cm

To Find :

Find the force exerted on the charge located at the 90 degree angle.​

Solution :

Force on 'B' due to charge at A (F₁) = $\frac{1}{4\pi E_0} * \frac{q^2}{r^2}$

                                                          = 9×10⁹ × $\frac{(10*10^-^8)^2}{(15*10^-2)^2}$

                                                          = 9×10⁹ × $\frac{100 * 10^-^1^6}{225 * 10^-^4}$

                                                          = 4 × 10⁻³ N

Force on 'B' due to charge at C (F₂) = $\frac{1}{4\pi E_0} * \frac{q^2}{r'^2}$

                                                           = 9×10⁹ × $\frac{(10*10^-^8)^2}{(20*10^-2)^2}$

                                                           = 9×10⁹ × $\frac{100 * 10^-^1^6}{400 * 10^-^4}$

                                                          = 2.25 × 10⁻³ N

Net Force acting on charge placed at B = $\sqrt{F_1^2 + F_2^2}$

                                                                  = $\sqrt{(4*10^-^3)^2 + (2.25*10^-^3)^2}$

                                                                  = 10⁻³ × $\sqrt{16 + 5.0625}$

                                                                  = 10⁻³ × $\sqrt{21.0625}$

                                                                 = 4.589 × 10⁻³ N

Therefore, the net force acting on the charge located at the 90 degree angle is 4.589 × 10⁻³ N.