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Physics Question on Electric Field

The electric field strength in N C −1N \,C\,^{-1} that is required to just prevent a water drop carrying a charge 1.6×10−19 C1.6 \times 10^{-19} \, C from falling under gravity is (g = 9.8 ms−2(g\, = \,9.8\, ms^{-2}, mass of water drop = 0.0016 g0.0016\,g)

A

9.8×10169.8 \times 10^{16}

B

9.8×10−169.8 \times 10^{-16}

C

9.8×10−139.8\times 10^{-13}

D

9.8×10139.8\times 10^{13}

Answer

9.8×10139.8\times 10^{13}

Explanation

Solution

Charge on drop =1.6×10−19C=1.6 \times 10^{-19} C Mass of the drop =0.0016 g=0.0016\, g =16×10−4 g=16 \times 10^{-4}\, g =16×10−7 kg=16 \times 10^{-7}\, kg g=9.8 m/s−2g=9.8\, m / s ^{-2} In balance position, mg=qEm g=q E where, EE is electric field strength. E=mgq=16×10−7×9.81.6×10−19E =\frac{m g}{q}=\frac{16 \times 10^{-7} \times 9.8}{1.6 \times 10^{-19}} =9.8×1013N/C=9.8 \times 10^{13} N / C