Two positively charged spheres of masses m_1 | vedclass

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(B) In the given situation,the forces acting on each charged sphere are:$(i)$ Gravitational force $(mg)$$(ii)$ Electrostatic repulsion force $F_e = \f

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$m_1 = m_2$

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(B) In the given situation,the forces acting on each charged sphere are:$(i)$ Gravitational force $(mg)$$(ii)$ Electrostatic repulsion force $F_e = \frac{k q_1 q_2}{r^2}$$(iii)$ Tension in the string $(T)$For equilibrium,we resolve the tension $T$ into horizontal and vertical components:$T \sin 	heta = F_e = \frac{k q_1 q_2}{r^2}$$T \cos 	heta = mg$Dividing the two equations,we get:$	an 	heta = \frac{F_e}{mg} = \frac{k q_1 q_2}{r^2 mg}$For sphere $1$:$	an 	heta = \frac{k q_1 q_2}{r^2 m_1 g}$For sphere $2$:$	an 	heta = \frac{k q_1 q_2}{r^2 m_2 g}$Since the angle $	heta$ is the same for both spheres,we equate the expressions:$\frac{k q_1 q_2}{r^2 m_1 g} = \frac{k q_1 q_2}{r^2 m_2 g}$This simplifies to:$m_1 = m_2$

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