Question 566: Electrostatics & Capacitance

Question

A charged oil drop is suspended in a uniform field of $$3 	imes {10^4}$$ $$v/m$$  so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $$ = 9.9 	imes {10^{ - 15}}\,\,kg$$ and $$g = 10\,m/{s^2}$$)

Accepted Answer

At equilibrium, electric force on drop balances weight of drop.
$$qE = mg \Rightarrow q$$
$$ = {{mg} \over E} = {{9.9 	imes {{10}^{ - 15}} 	imes 10} \over {3 	imes {{10}^4}}}$$
$$ = 3.3 	imes {10^{ - 18}}C$$

Answer

$$1.6 	imes {10^{ - 18}}\,C$$

Answer

$$3.2 	imes {10^{ - 18}}\,C$$

Answer

$$3.3 	imes {10^{ - 18}}\,C$$

Answer

$$4.8 	imes {10^{ - 18}}\,C$$

Electrostatics & Capacitance

A charged oil drop is suspended in a uniform field of $3 \times {10^4} v/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9 \times {10^{ - 15}}\,\,kg and g = 10\,m/{s^2}$)

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