Question

a oil drop of mass m is suspended in equilibrium b/w two horizontal conducting plates each of area A m2 ..and having charge +q and -q C .find charge on drop​

Answer 1

Answer: A€mg/q

Explanation:

Since, the electric field b/w two conducting plates each of area A

=q/A×ebsilnot

and we also know that F=qE

F=Q×(q/A×ebsilnot). where Q is charge on oil drop

F=Qq/A×ebsilnot

Now,since from the following figure we can see that

mg=F

mgA×ebsilnot/q=Q

Soi, the charge iisQ=A€mg/q

​

PLZ MARK IT BRAINLIEST

Answer 2

The charge on drop​ is = ( A×m×g×ε0 ) / q   C

Given: The mass of an oil drop is 'm'

The area of conducting plates is A m² and has charge +q C and -q C

To Find: Charge on the drop​ of oil.

Solution:

Let the charge on the oil drop be 'Q'.

The gravitational force on the drop ( Fg ) = mg

Electric force on the drop ( Fe ) = QE                 [ where E = Electric field]

The electric field ( E ) = q / ( Aε0 )        

  [ where q = charge on plate, ε0 = permittivity of free space ]

Thus, Electric force on the drop ( Fe ) = Q × ( q / ( Aε0 )      

In equilibrium, Fg = Fe

        ⇒ mg =  Q × ( q / ( Aε0 )

        ⇒ Q = ( A×m×g×ε0 ) / q

Hence, the charge on drop​ is =  ( A×m×g×ε0 ) / q   C

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