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Physics Question on Electric Charge

A charged oil drop is suspended in a uniform field of 3?104 V/m3 ? 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 ? 10^{-15}\, kg and g=10 m/s2g = 10\, m/s^2)

A

3.3?10−18 C3.3 ? 10^{-18}\, C

B

3.2?10−18 C3.2 ? 10^{-18}\, C

C

1.6?10−18 C1.6 ? 10^{-18}\, C

D

4.8?10−18 C4.8 ? 10^{-18}\, C.

Answer

3.3?10−18 C3.3 ? 10^{-18}\, C

Explanation

Solution

Since ball is hanging in equilibrium, force by gravity is balanced by electric force. qE=mgqE = mg ⇒q=m×gE\Rightarrow q = \frac{m\times g}{E} ⇒9.9×10−15×103×104\Rightarrow \frac{9.9\times 10^{-15}\times 10}{3\times 10^{4}} ∴q=3.3×10−18 C\therefore q = 3.3\times 10^{-18}\,C