Question

1, An oil drop mass carrying charge - Q is held stationary in the gravitational field of the earth. What is the magnitude and direction of the electrostatic field required for this purpose? Ans: E = mg / Q, downward

Answer 1

Answer:

Given:

Charge in oil drop = -Q

Mass be m

Gravity be g

Electric Field Intensity is E

To find:

Electric Field intensity such that the Oil drop will float in air and remain stationary.

Concept:

For the oil drop to be stationary or floating , the oil drop must be in equilibrium under equal and opposite forces.

The forces have to be :

• Gravitational Force
• Electrostatic Force

Since the charge is negative , the electric Field line should direct downwards such that It experience force upwards.

Similarly Gravity will pull the charge downwards.

Calculation:

Electrostatic Force = Gravitational Force

$= > E \times Q \: = m \times g$

$= > E \: = \dfrac{m \times g}{Q}$

$= > E \: = \dfrac{mg}{Q}$

So final answer :

$\boxed{ \red{ \huge{ \sf{ \bold{E \: = \dfrac{mg}{Q} (\downarrow \downarrow)}}}}}$

Answer 2

SOLUTION

Given,

• Mass of the Oil Drop = M

• Charge = - Q

Let $\sf F_G \ and \ F_E$ be the gravitational force and electric force acting on the oil drop respectively

If the oil drop is at rest,the forces acting on the drop are in equilibrium

Gravitational Force

$\sf F_G = Mg-------(1)$

Electric Force

$\sf E =- \dfrac{F_E}{Q} \\ \\ \longrightarrow F_E = - EQ---------(2)$

Since,the force are in equilibrium. Both the forces are equal in magnitude and cancel out eachother

Thus,from (1) and (2) :

$\sf -EQ =Mg \\ \\ \longrightarrow \boxed{\boxed{\sf E=- \dfrac{Mg}{Q} }}$

Direction of the electric field is downwards ( Negative Y axis)