Two metal plates (A, B) are kept horizontally | vedclass

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(B) The force of gravity on the oil droplet is balanced by the electric force acting upwards. The droplet must be attracted towards the top plate $A$.

Vedclass · Accepted Answer

$580 	ext{ V}$ and $100 	ext{ V}$

Vedclass · Answer

(B) The force of gravity on the oil droplet is balanced by the electric force acting upwards. The droplet must be attracted towards the top plate $A$. The force of gravity is $F_g = mg = (ho \cdot \frac{4}{3}\pi r^3)g$.The electric force is $F_e = qE = q(\frac{V_A - V_B}{d})$.Setting $F_g = F_e$, we have $ho \cdot \frac{4}{3}\pi r^3 g = q \frac{\Delta V}{d}$.Given $ho = 1500 	ext{ kg/m}^3$, $r = 10^{-3} 	ext{ m}$, $g = 10 	ext{ m/s}^2$, $d = \frac{0.12}{\pi} 	ext{ m}$, $q = 5 	imes 10^{-9} 	ext{ C}$.Substituting values: $1500 	imes \frac{4}{3} 	imes \pi 	imes (10^{-3})^3 	imes 10 = 5 	imes 10^{-9} 	imes \frac{\Delta V}{(0.12/\pi)}$.$20 	imes 10^{-6} 	imes \pi = 5 	imes 10^{-9} 	imes \frac{\Delta V 	imes \pi}{0.12}$.Solving for $\Delta V$: $\Delta V = \frac{20 	imes 10^{-6} 	imes 0.12}{5 	imes 10^{-9}} = 480 	ext{ V}$.Since $V_A - V_B = 480 	ext{ V}$, only option $(B)$ $580 	ext{ V} - 100 	ext{ V} = 480 	ext{ V}$ satisfies this condition.

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