Electrostatics Part 3

Concept:At equilibrium the net force on the bob is zero, so the tension must balance the vector sum of the weight and the electric force.The weight $mg$ acts vertically downward.The electric force $qE$ acts horizontally (since the field is horizontal and the bob is charged).Explanation:Let the string make an angle $\theta$ with the vertical in the equilibrium position.Resolve the tension into components:Vertical component: $T \cos \theta = mg$.Horizontal component: $T \sin \theta = qE$.Square and add the two equations:$T^2(\sin^2 \theta + \cos^2 \theta) = (mg)^2 + (qE)^2$.Since $\sin^2 \theta + \cos^2 \theta = 1$,$T^2 = m^2 g^2 + q^2 E^2$.Therefore, $T = \sqrt{m^2 g^2 + q^2 E^2}$.Note that $mg + qE$ or $mg - qE$ would only be valid if the two forces acted along the same line, which is not the case here.Answer:Option B: $\sqrt{m^2 g^2 + q^2 E^2}$