Solution
The disc leaves the dome when the Normal force $N$ becomes zero. The forces acting on the disc are gravity ($mg$) acting downwards and the electrostatic force ($qE$) acting horizontally.
Given $\theta = \sin^{-1}(3/5)$, we have $\sin\theta = 0.6$ and $\cos\theta = 0.8$.
Radial Force Equation (at separation, N=0):
$$mg \cos\theta – qE \sin\theta = \frac{mv^2}{R} \quad \dots(1)$$Work-Energy Theorem:
Work done by gravity + Work done by electric field = Change in Kinetic Energy
$$mgR(1 – \cos\theta) + qE(R \sin\theta) = \frac{1}{2}mv^2 \quad \dots(2)$$From (1), $mv^2/R = mg \cos\theta – qE \sin\theta$. Substituting this into (2) multiplied by 2/R:
$$2mg(1 – \cos\theta) + 2qE \sin\theta = mg \cos\theta – qE \sin\theta$$ $$2mg – 2mg \cos\theta + 2qE \sin\theta = mg \cos\theta – qE \sin\theta$$ $$3qE \sin\theta = 3mg \cos\theta – 2mg$$ $$qE (3 \sin\theta) = mg (3 \cos\theta – 2)$$Substituting values ($\sin\theta=0.6, \cos\theta=0.8$):
$$qE (3 \times 0.6) = mg (3 \times 0.8 – 2)$$ $$1.8 qE = mg (2.4 – 2) = 0.4 mg$$ $$\frac{mg}{qE} = \frac{1.8}{0.4} = \frac{18}{4} = \frac{9}{2}$$Answer: (c) 9/2