Home
Class 12
PHYSICS
In Millikan's oil drop of mass 16xx10^(-...

In Millikan's oil drop of mass `16xx10^(-6)`kg is balanced by an electric field of `10^6V//m`. The charge in coulomb on the drop is: (assuming `g=10 m//s^2`)

A

`6.2xx10^(-11)`

B

`16xx10^(-9)`

C

`16xx10^(-11)`

D

`1.6xx10^(-13)`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

In Millikan's oil drop experiment, an oil drop mass 60 xx 10^(-6) kg is balanced by an electric field of 10^(6) V//m . The charge in coulomb on the drop, assuming g = 10 m//s^(2) is

If an oil drop of weight 3.2xx10^(-13) N is balanced in an electric field of 5xx10^(5) Vm^(-1) , find the charge on the oil drop.

In a Millikan's oil drop experiment, an oil drop of mass 4.9xx10^(-14)kg is balanced by a potential difference of 4000 V between two plates which are 8 mm apart. Find the number of elementary charges on the drop.

An oil drop of mass m and charge +q is balanced in vaccum by a uniform electric field of intensity E . the direction of this field should be

A charged ball of mass 8.4xx10^(16) kg is found to remain suspended in a uniform electric field of 2xx10^(4) Vm^(-1) . Calculate the charge on the ball. Given g = 10 m//s^(2)

In 1909, Robert Millikan war the firest to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of opposite charged plates. The drops were pbserved with a magnitue eyepiece, and the electric field was adjusted so that upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of 1.6xx10^(-19)C- the charge of the electron. For this, he won the Nobel prize. If a drop of mass 1.08xx10^(-14)kg remains stationary in an electric field of 1.68xx10^(5)NC^(-1) , then teh charge of this drop is

A pendulum bob of mass 30.7 xx 10^(-6)kg and carrying a chargee 2 xx 10^(-8)C is at rest in a horizontal uniform electric field of 20000V//m . The tension in the thread of the pendulum is (g = 9.8 m//s^(2))

A drop of 10^(-6) kg water carries 10^(-6) C charge. What electric field should be applied to balance its weight (assume g = 10 ms^(-2) )

A particlem of mass 2 xx 10^(-3) kg, charge 4 xx 10^(-3)C enters in an electric field of 5 V//m , then its kinetic energy after 10 s is