Electric potential difference question

[jaejoon89]

Two parallel horizontal plates are 0.4 cm apart in air. An oil droplet of m = 4.9*10^-17 kg is introduced between the plates. If the drop has 2 electronic charges and if there is no air buoyancy, you can hold the droplet motionless between the plates if you keep the potential difference between them at...?



Do I use delta V = -int[E*dl] from a to b? How would I incorporate the mass?

 

[cepheid]

For your first question, yes, that is the definition of electric potential.

For your second question, the mass determines how much weight the electric field has to counteract.

 

[baba89]

Yes, you can use the equation delta V = -int[E*dl] from a to b to determine the potential difference between the plates. The electric field (E) can be calculated using the formula E = V/d, where V is the voltage and d is the distance between the plates. In this case, the distance is given as 0.4 cm.

To incorporate the mass, you can use the equation F = qE, where F is the force, q is the charge, and E is the electric field. Since the droplet has 2 electronic charges, the force acting on it would be 2qE. This force must be balanced by the gravitational force, which is given by mg, where m is the mass of the droplet and g is the acceleration due to gravity.

Setting these two forces equal to each other, we get 2qE = mg. Rearranging this equation, we can solve for the potential difference (V) as V = (mg)/(2q). Substituting in the given values, we get V = (4.9*10^-17 kg * 9.8 m/s^2)/(2*1.6*10^-19 C) = 15.3 V. Therefore, to hold the droplet motionless between the plates, you would need to keep the potential difference between them at 15.3 V.