JEE MAIN - Physics (2004 - No. 14)
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 $$g = 10\,m/{s^2}$$)
$$1.6 \times {10^{ - 18}}\,C$$
$$3.2 \times {10^{ - 18}}\,C$$
$$3.3 \times {10^{ - 18}}\,C$$
$$4.8 \times {10^{ - 18}}\,C$$
Explanation
At equilibrium, electric force on drop balances weight of drop.
$$qE = mg \Rightarrow q$$
$$ = {{mg} \over E} = {{9.9 \times {{10}^{ - 15}} \times 10} \over {3 \times {{10}^4}}}$$
$$ = 3.3 \times {10^{ - 18}}C$$
$$qE = mg \Rightarrow q$$
$$ = {{mg} \over E} = {{9.9 \times {{10}^{ - 15}} \times 10} \over {3 \times {{10}^4}}}$$
$$ = 3.3 \times {10^{ - 18}}C$$
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