Question

A charged oil drop is suspended in a uniform field of $3 × 10^4 \,V/m$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge $= 9.9 × 10^{−15}\, kg$ and $g = 10\, m/s^2$)

• A. $3.3 × 10^{−18}\, C$
• B. $3.2 × 10^{−18}\, C$
• C. $1.6 × 10^{−18}\, C$
• D. $4.8 × 10^{−18}\, C$.

Answer 1

Answer: (A)

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