Assignment 3: (Spring 2015)
TOTAL MARKS: 25
Due Date: 25/01/2016
DON’T MISS THESE Important instructions:
To solve this assignment, you should have good command over first 29 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
Two particles having charges q1 = 0.500 nC and q2 = 8.00 nC are separated by a distance of 1.20 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero? Marks 8
Question # 2
If the electric field in a region of space is zero, can you conclude that no electric charges are in that region? Explain. Marks 5
Question # 3
If the total charge inside a closed surface is known but the distribution of the charge is unspecified, can you use Gauss's law to find the electric field? Explain. Marks 5
Question # 4
Assume that a deuteron has a mass of 3.34 × 10-27Kg and a charge of +e. It travels in a circular path with a radius 6.96mm in a magnetic field with magnitude 2.50T. Value of charge on electron is 1.6 × 10-19C.
Find the speed of the deuteron. Marks 7
.+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)+ Click Here to Search (Looking For something at vustudents.ning.com?) + Click Here To Join (Our facebook study Group)
I think wrong....
Kia hum r ko mm sy Meter main convert nahi kerain gy????
aur KG ko grams main????
Complete assignment day do yaar koi :/
First charge = q1=0.500 nC
Second charge = q2 = 8.00 nC
Distance between the charge r=d=1.20m
The electric field generated by a point charge is
E = (Q/r^2)(1/((4)(pi)(e)))
Where e is the permittivity of free space.
You can set up the two fields and make them equal to each other:
(Q1/r^2)(1/((4)(pi)(e))) = (Q2/(1.2-r)^2)(1/((4)(pi)(e)))
We need to solve for r, where r is the distance from Q1.
the (1/((4)(pi)(e))) cancels on both sides to get:
Q1/r^2 = Q2/(1.2-r)^2 plug in values for Q1 and Q2:
.5/r^2 = 8/(1.2-r)^2 (notice the units for charge doesn't matter since they will cancel out as long as they are in the same units) solve for r:
(1.2-r)^2 = 16r^2
1.2-r = 4r
1.2 = 5r
r = 1.2/5 = .24m
so The electric fields are equal when you are .24m away from the .5nC charge and since the electric fields act in opposite directions (both are positively charged particles) the field will equal zero there.
Not necessarily. If the magnetic field is parallel or anti-parallel to the velocity of the charged particle, then the particle will experience no magnetic force. There may also be an electric force acting on the particle such that! F q = q (! E+! V q!! B) = 0.
M= 3.34 × 10-27Kg
+e=1.6 × 10-19C.
phy101 assignment solution 2016