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Physics Question on coulombs law

A charged oil drop is suspended in uniform field of 3×104 V/m3 \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 =9.9×10−15 kg= 9.9 \times 10^{-15}\, kg and g=10 m/s2)g = 10\, m/s^2)

A

3.3×10−18 C3.3 \times 10^{-18}\, C

B

3.2×10−18 C3.2 \times 10^{-18}\, C

C

1.6×10−18 C1.6 \times 10^{-18}\, C

D

4.8×10−18 C4.8 \times 10^{-18}\, C

Answer

3.3×10−18 C3.3 \times 10^{-18}\, C

Explanation

Solution

Given : mass o f charged drop m=9.9×10−15 kgm=9.9\times 10^{-15}\,kg Electric field E=3×104V/mE=3\times 10^{4} V/m In steady state. Electric force on a drop = weight of a drop ∴qE=mg\therefore qE=mg or q=mgEq=\frac{mg}{E} =9.9×10−15×103×104=\frac{9.9\times10^{-15}\times10}{3\times 10^{4}} =3.3×10−18 C=3.3\times 10^{-18}\,C