Question

An oil drop of mass M is found floating freely between the plates of a parallel plate condenser, the plates being
horizontal and the lower plate carrying a charge + Q and charge on upper plate is Q. The area of each plate
Is A and the distance of separation between them is D. Then calculate the charge on the oil drop (the acceleration
due to gravity is g) (Electric field between the plates of a parallel plate condenser E =-
ENEO
Workout the magnitude and direction of field at noint P when a charge of 2ur exneriences an alactrical force​

Answer 1

Given that,

Mass = M

Charge on the plate = Q

Area = A

Distance = D

Acceleration due to gravity = g

Electric field = E

Let the charge on the oil drop is q.

We need to calculate the charge on the oil drop

Using formula of force

$F=mg$

$qE=mg$

Put the value into the formula

$qE=Mg$

Where, q = charge

E = electric field

M = mass

g = acceleration due to gravity

Put the value into the formula

$q\dfrac{Q}{A\epsilon_{0}}=Mg$

$q=\dfrac{A\epsilon_{0}Mg}{Q}$

Hence, The charge on the oil drop is $\dfrac{A\epsilon_{0}Mg}{Q}$

Answer 2

Answer:

Step-by-step explanation:

Given that,

Mass = M

Charge on the plate = Q

Area = A

Distance = D

Acceleration due to gravity = g

Electric field = E

Let the charge on the oil drop is q.

We need to calculate the charge on the oil drop

Using formula of force

Put the value into the formula

Where, q = charge

E = electric field

M = mass

g = acceleration due to gravity

Put the value into the formula

Hence, The charge on the oil drop is