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Assignment # 3 of Physics PHY101 has been uploaded. Last date to submit your assignment solution is 25th January -2016.

Assignment 3: (Spring 2015)

PHYSICS (PHY101)

TOTAL MARKS: 25

Due Date: 25/01/2016

 

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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


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Question#1..... Answer

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.

baqi 2 ans..? 

Question # 2 . Answer.....

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 .

baqi k b answer day do brother

Still Waiting Please koi to bata do is Assignment ka Solution....

Q No 3 and 4 is required please? any one help?

please solution help phy101 

Please Discuss here about this assignment.Thanks
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phy101=3

Question#1..... Answer
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.

Question # 2 . Answer.....
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 .

Qno3 ka ans shre kr dain plz

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

Ans:

 

     M= 3.34 × 10-27Kg

     +e=1.6 × 10-19C.

    

     

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