Question

What voltage is needed to balance an oil drop carrying 5 electrons when located between the plates of a capacitor 5 mm apart? The mass of the drop is $3.12\times {10}^{-16}kg.$

• A. 15.5 V
• B. 17.2 V
• C. 19.1 V
• D. 21.7 V

Answer 1

Answer: (C)

19.1 V

given
mass(m) = $3.12\times {10}^{-16}$ kg
charge (q) = $5\times 1.602\times {10}^{-19}$ coulomb
distance(d) = $5\times {10}^{-3}$ m
and we know that g = 9.8 $m.s^{-2}$
substitute these values in the following equation and solve

V = $\frac{(3.12\times {10}^{-16})\times \left(9.8\right)\times (5\times {10}^{-3})}{5\times 1.602\times {10}^{-19}}=19.1V$

So, the answer is option (C).