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Physics Question on coulombs law

A charged oil drop is suspended in a uniform electric 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 drop is 9.9×10−15 kg9.9 \times 10^{-15} \,kg and g=10 ms−2g=10 \,ms ^{-2} )

A

3.3×1018C3.3\times {{10}^{18}}C

B

3.2×10−18C3.2\times {{10}^{-18}}C

C

1.6×10−18C1.6\times {{10}^{-18}}C

D

4.8×10−18C4.8\times {{10}^{-18}}C

Answer

3.3×1018C3.3\times {{10}^{18}}C

Explanation

Solution

Because the drop neither falls nor rises, so it remains stationary, then
For equilibrium position, the electric farce qEq E will be equal to the weight of the drop.
So, qE=mgq=mgEq E=m g q=\frac{m g}{E}
=(9.9×10−15)×103×104=\frac{\left(9.9 \times 10^{-15}\right) \times 10}{3 \times 10^{4}}
=3.3×10−18C=3.3 \times 10^{-18} C