Notes 2 for June 18 prepared by Melanie Smith
Shusaku discussed certain rules and regulations for how out discussion section will work.
Everyone responsible for creating problems for the next day’s lecture should send them to
the compiler of the day, who will then email the completed problem sheet to the entire
class as soon as possible.
Shusaku also informed us about the laboratory procedures for tomorrow. Everyone should
have read the lab (we are doing the first lab in the manual) and have a general idea of
what we will be doing. Do your best to answer the pre-lab questions. There will be a 10
minute quiz tomorrow, consisting of approximately 3 questions, but this is a practice quiz,
and will not count for a grade. The next lab will contain a quiz that counts for credit, with
3 questions on the new lab, and two on the previous lab.
Solutions to the Conceptual Problems:
1)
Describe the motion of a proton after it is released from rest in a uniform electric field .
A proton released from rest will move away from the positive source in the direction of the
negative source.
Describe the changes (if any) in its kinetic energy and the electric potential energy
associated with the proton.
The Kinetic Energy will increase because the proton will accelerate from rest. The Potential
Energy (V) will decrease because the two like charges a
re farther apart.
As r increases, v decreases
As KE increases, PE decreases
V = kq/r
E = F/q and E = K(Q/r2)
F = qE
If F is constant, E is constant, and since F = ma, the proton will move at a constant rate.
We skipped going over this problem, as we have not gotten to it in the lecture notes yet.
Electrical force and gravitational forces are two non-contact forces. In considering the fact
that Coulomb’s Law equation for electrical force shows a strong resemblance to Newton’s
equation for universal gravitation, discuss both the similarities and differences between
electrical and gravitational forces.
Similarities:
- Inverse square Law
Differences:
electrical field is stronger
gravity --> mass, electrical --> charge
gravity --> attractive, electrical --> attractive and repulsive
It is observed that Balloon A is charged negatively. Balloon B exerts a repulsive affect upon
Balloon A. Would the electric field vector by Balloon B be directed towards or away from
Balloon B? Explain.
Answer: Electrical field vector is pointing towards the RIGHT
Explanation: If the charge is negative, the electric field vector points towards the opposite
direction as the force vector. When there is a positive charge, the electric field and force
vector both point towards the same direction. Also, the vector always goes from a positive
to a negative charge. Therefore, the arrows all point IN towards balloon B (as well as in
towards Balloon A, as both are negative charges) and the electric field vector ultimately
points towards the right.
* Unfortunately, we did not get a chance to go over the Quantitative Problems, as we ran
out of time. Tomorrow’s discussion will review these solutions. :)