Question

The voltage required to balance an oil drop carrying 10 electrons between the plates of a capacitor which are 10 mm apart, is (Given mass of the oil drop$=3.2\times {10}^{-15}kg,e=1.6\times {10}^{-19}C)$ :

• A. 16 V
• B. 160 V
• C. 196 V
• D. 19.6V

Answer 1

The correct option is C 196 V

Let the voltage required by $V_0$
For equilibrium,
Weight $=$ Upward force
$\Rightarrow 3.2\times {10}^{-15}\times 9.8=16\times {10}^{-19+2}\times V_0$
$\Rightarrow V_0=196volts$

So, the answer is option (C).