Solving for Electric Field Strength & Direction of Charged Sphere

AI Thread Summary
The discussion focuses on calculating the electric field strength required to balance the weight of a charged plastic sphere. The sphere weighs 1.0g and has a charge of -3nC, leading to the formula E = mg/q, resulting in an electric field strength of 3x10^6 N/C. The net force equation indicates that the electric force must equal the gravitational force for balance. The direction of the electric field is determined to be opposite to the gravitational force, as the sphere is negatively charged. Thus, the electric field must point upward to counteract the downward force of gravity.
Lydia22
Messages
33
Reaction score
0

Homework Statement



What are the strength and direction of an electric field that will balance the weight of a 1.0g plastic sphere that has been charged t0 -3nC ?


Homework Equations


E= F/q



The Attempt at a Solution



Newton's laws on the charged sphere

ƩFnet(y) = F -mg=0
F=mg\
F=qE
E = mg/q
=[(0.003kg)(9.8m/s^2)]/(3.0x10^-9C)
=3x10^6 N/c
how do i conclude on the direction?
 
Physics news on Phys.org
Hi Lydia22! :smile:

You have already used what the direction must be.
So you only have to make your conclusion explicit.

In your first line you have ƩFnet(y) = F -mg=0, which is the condition for the sphere to be in balance.
This implies that the electric force is equal and opposite to the force of gravity.
Since q is positive, that also holds true for the electric field.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top