What is the charge needed to suspend an oil drop in an electric field?

[marshall4]

What charge is required to suspend a 1.7*10^-13kg oil drop in an electric field of 2.5*10^5 N/C?

 

[himanshu121]

Draw the Free Body Diagram
Apply $$\Sigma F$$ = $$\vec F_m_a_g + \vec F_w_t$$ = 0
As in case of equilibrium

Also apply the charge quantization

 

[M98Ranger]

To solve this problem, we can use the equation F = qE, where F is the force on the oil drop, q is the charge on the oil drop, and E is the electric field strength.

We know that the force of gravity on the oil drop is equal to its weight, which can be calculated using F = mg, where m is the mass of the oil drop and g is the acceleration due to gravity (9.8 m/s^2).

Since the oil drop is suspended, the force of gravity must be equal to the force of the electric field, so we can set these two equations equal to each other:

qE = mg

We can rearrange this equation to solve for q:

q = mg/E

Plugging in the given values, we get:

q = (1.7*10^-13kg)(9.8 m/s^2)/(2.5*10^5 N/C)

Simplifying, we get:

q = 6.66*10^-22 C

Therefore, a charge of 6.66*10^-22 C is required to suspend the 1.7*10^-13kg oil drop in an electric field of 2.5*10^5 N/C.