Question

A charged oil drop of mass 10 µg and charge 10 nC is suspended between two parallel horizontal charged plates. The electric field between the plates is

4 N/C


6 N/C


10 N/C


8 N/C

Answer 1

Given:

A charged oil drop of mass 10 µg and charge 10 nC is suspended between two parallel horizontal charged plates.

To find:

Electric field between plates ?

Calculation:

Since the charged oil drop is in equilibrium, the weight of the charged drop is being balanced by the electrostatic force.

$\sf \: E \times q = mg$

• Putting all values in SI UNIT:

$\sf \implies\: E \times (10 \times {10}^{ - 9} )= (10 \times {10}^{ - 9} ) \times 10$

• Cancelling 10 × 10^(-9) on both sides:

$\sf \implies\: E = 10 \: N/C$

So, field intensity inside plates is 10 N/C.