1. 

   Ϻαᴊοᴙα
   July 27th, 2010 - 19:30

   The sum of the forces in the vertical (y) direction must equal 0, as the charge is not accelerating. The two forces acting on the charge are the force of the electric field and the force of gravity. We also know that the particle is hovering. If the charge were positive, the electric field force and the force of gravity would both be acting downwards, leaving no way for the sum of the two forces to equal zero if they are both negative. If the charge is negative, they it will want to go upwards, against the electric field, and gravity would be trying to pull it down. This is the only way for the particle to have the forces acting on it equal zero. With this knowledge, we can come up with the expression:

   ∑Fy = Fe – mg = 0
   (drawing a free body diagram always helps with this step)

   Now, we know that Fe = Eq, so we can plug in our values and solve for q:
   (120N/C)q – (0.00016kg)(9.8m/s^2) = 0
   q = 1.31×10^-5 C
2. 

   David
   July 27th, 2010 - 19:44

   Yes very easy
   The force F that the electric field exerts on the charge q is acting upwards and the weight of the charged drop is acting downwards due to the gravitational pull
   The two forces must balance for the drop to remain stationary F= qE = mg

   I’m sure you can do the maths now to get your answer.

   The whole thing is related to Millikan’s famous oil drop experiment to find the charge on an electron

Leave a comment

Name (required)

Mail (will not be published) (required)

Website

( Cancel )

CAPTCHA Code *