Assignment
3: (Spring 2015)
PHYSICS (PHY101)
TOTAL MARKS: 25
Due Date: 27/07/2015
DON’T
MISS THESE Important instructions:
To solve this assignment, you should have good command over
first 27 lectures.
Upload assignments (Microsoft word) properly through LMS, (No
Assignment will be accepted through
email).
Write your ID on the top of your solution file.
All students are directed to use the font and style of text
as is used in this document.
Don’t use colorful back grounds in your solution files.
Use Math Type or Equation Editor etc for mathematical
symbols.
Write to the point and avoid from unnecessary explanation. Don’t
copy/paste from internet or any other source.
This is not a group assignment, it is an individual
assignment so be careful and avoid copying others’ work. If some assignment is
found to be copy of some other, both will be awarded zero marks. It also
suggests you to keep your assignment safe from others. No excuse will be
accepted by anyone if found to be copying or letting others copy.
Don’t wait for the last date to
submit your assignment.
Question
# 1
A
particle with a charge of 0.17coulombs is in a uniform electric field of
strength 55 N/C as shown below. An external force pushes the charge 0.12 meters
directly against the field. What is the change in electric potential energy?
Marks 8
Question # 2
A) Where does an electron go when it
is free to move: to a location of higher or lower potential? Explain it.
B) A student of Physics calculated a
negative potential difference between two points. Is that possible? Explain it
by giving a solid reason. Marks
3 +3 = 6
Question # 3
What is the electric potential at the location of
the test charge as shown in the figure?
Marks 5
Question # 4
What is the potential difference between the plates
shown in the following figure? Marks 6
Solution complete
Question:1
Ans.
q= .17 C
∆x= .12 m
E= 55 N/C
In order to find Change in Electric Potential Energy = ∆Ue = ?
We need electric potential = V = ?
∆U e = q∆V
∆V = -E∆x = -(55 N/C)(0. 12 m) = – 6.6 Nm /C = -6.6 J/C
Now to find:
∆Ue = q∆V = (0.17C)(-6.6J/C)= 1.122 J
∆x= .12 m
E= 55 N/C
In order to find Change in Electric Potential Energy = ∆Ue = ?
We need electric potential = V = ?
∆U e = q∆V
∆V = -E∆x = -(55 N/C)(0. 12 m) = – 6.6 Nm /C = -6.6 J/C
Now to find:
∆Ue = q∆V = (0.17C)(-6.6J/C)= 1.122 J
Question:2
(a)
Ans.
As electron has negative charge, it will always move towards
the pole containing positive charge, Irrespective of the fact that, which potential
you denote as higher and which lower.
(b)
Ans.
Negative potential means potential in opposite direction.
For example if we connect the negative terminal of a battery to ground than our
meter will calculate positive potential difference, but if we connect it other
way round than our meter will calculate negative potential difference.
Question:3
Question:4
Potential difference between plates
Vb = 1.0 volts
Va = -0.5 volts
∆V = Vb – Va = 1.0- (-0.5)
∆V = 1.5 Volts
Post a Comment
0 comments
Dear readers, after reading the Content please ask for advice and to provide constructive feedback Please Write Relevant Comment with Polite Language.Your comments inspired me to continue blogging. Your opinion much more valuable to me. Thank you.