In Millikan's oil drop experiment, an oil drop of radius and charge Q is held in equilibrium between the plates of a charged parallel plate capacitor when the potential difference is V. To keep a drop of radius 2 r and charge 2 Q in equilibrium between the plates, the potential difference V prime required is

Question

In Millikan's oil drop experiment, an oil drop of radius and charge Q is held in equilibrium between the plates of a charged parallel plate capacitor when the potential difference is V. To keep a drop of radius 2 r and charge 2 Q in equilibrium between the plates, the potential difference V prime required is (A) 8 V (B) 4 V (C) 2 V (D) V

Tardigrade Admin · Accepted Answer

At equilibrium, Q ⋅ E=m g=((4/3) π r3) ρ ⋅ g (Q ⋅ V/d)=((4/3) π r3) ρ ⋅ g ⇒ V=((4/3) π r3) (ρ ⋅ g ⋅ d/Q) If charge becomes 2 Q and radius becomes 2 r, then V=[(4/3) π(2 r)3] (ρ ⋅ g ⋅ d/(2 Q)) ⇒ V=((23/2)) V=4 V

$8 V$

$4 V$

$2 V$

$V$

8V

4V

2V

V

At equilibrium, Q⋅E=mg=(34​πr3)ρ⋅g dQ⋅V​=(34​πr3)ρ⋅g⇒V=(34​πr3)Qρ⋅g⋅d​ If charge becomes 2Q and radius becomes 2r, then V=[34​π(2r)3](2Q)ρ⋅g⋅d​⇒V=(223​)V=4V

$8 V$

$4 V$

$2 V$

$V$