Question

Three equal charges of 10° C respectively, each Yocated at the corners of a right triangle whose sides are 3 cm, 4 cm and 5 cm respectively. Find the force exerted on the charge located at the 90° angle. (E = 8.85 x 10-12 SI unit)​

Answer 1

Given, Three equal charges of 10 coulombs respectively, each located at the corners of a right-angle triangle whose sides are 3cm, 4cm, and 5cm respectively. The given value of E = 8.85 × $10^{-12}$ SI unit.

To Find, What is the force exerted on the charge located at the 90° angle?

Solution, From Coulomb's law we know that,

F = k$\frac{q_{1}q_{2} }{r^{2} }$

  = $\frac{1}{4\pi e_{0} }$$\frac{q_{1}q_{2} }{r^{2}}$

Let us consider the charges are located at the vertexes of triangle ABC and angle ABC is equal to 90°.

Now the force along AB = 100k/4² Newton

The force along BC = 100k/3² Newton

Now the force on the charge located at point B will be,

F = $\sqrt{AB^{2}+BC^{2} + 2AB.BC Cos90^{0}$

F= $\sqrt{AB^{2}+BC^{2}+ 2AB.BC.0 }$     [ cos 90° = 0 ]

F =$\sqrt{AB^{2}+BC^{2} }$

F=$\sqrt{(100k/4^{2}) ^{2} +(100k/3^{2})^{2}$  

F= $\sqrt{(10000K^{2}/16^{2})+ (10000K^{2} /9^{2}) }$

Where, k = 1/4πe₀ = 8.99 × 10⁹ N.m²/c²

F = $\sqrt{(10000/256+10000/81)K^{2} }$

F = $\sqrt{(39.06+123.46)(8.99*10^{9} )^{2} }$

F = $\sqrt{(162.52)(80.82)(10^{18}) }$

F= $\sqrt{(13,134.86)(10^{18}) }$

F= 114.61×10⁹ Newton or N

Hence, the force exerted on the charge located at the angle of 90° will be 114.61×10⁹ N.