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PHY101 Physics Assignment No 03 Solution & Discussion Due Date:11-01-2013

Physics (PHY101) VU of Pakistan
Assignment 3: (Fall 2012)
PHYSICS (PHY101)
TOTAL MARKS: 35
Due Date: 11/01/2013
DON’T MISS THESE Important instructions:
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Don’t wait for the last date to submit your assignment.
Question # 1
(a)What is the electric field at the location of the test charge shown below? And what is the
direction of electric field? Marks = 5
Physics (PHY101) VU of Pakistan
(b) What are the magnitude and direction of the electric field at point A shown below? Marks = 6
Question # 2
Find the force on the electron in figure shown below and also determine the acceleration of
electron? Mention the direction of force and acceleration as well.
Note that mass of electron is 9.1×10-31kg, and the value of charge on electron is (-1.6×10-19 C)
Marks = 7
Question # 3
Is it true to say that electric field lines are the same as electric field vectors. If yes give an
example if not explain it. Marks = 6
Question # 4
Suppose you increased the amount of the charge of a test charge, and the force exerted on it
changed. This means the field you were assessing must have changed. Marks = 5
Question # 5
(a) It takes 10.0 J of work by the right-hand wand to separate the charges as shown in figure.
What is the change in potential energy? Marks = 3
Physics (PHY101) VU of Pakistan
(b) Is it true to say that electric potential energy and electric potential are the same?
Either yes or no explain in each case. Marks = 3
<Wish you good luck>>>>>>>>>>>>

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Replies to This Discussion

Q1a

F=1.2N

Q=2.5µc=2.5*10-6 Coulombs

Use the formula E=F/q

E=1.2/2.5*10-6

E=0.48*106

E=4.8*105N/c

E=480000N/C

Q1b

K=9.0*109 Nm2/c2

q=12.5 C

r=11.5cm=11.5*10-2m

Use the formula

E=K*q/r2

E=9.0*109 *12.5/(11.5*10-2) Nm2/C2 *C/m2

E=9*12.5/11.5*11.5*109*104N/C

E=0.85066*1013N/C

E=85.066*1011N/C

Q-2

E=2.0*10-6N/C

Mass=m=9.1*10-31kg

Charge=q=-1.6*10-19C

By using formula

F=E*q

F=2.0*10-6*(-1.6*10-19)(N/C)(C)

F=3.2*10-25N

For acceleration

Acceleration(a)=F/m

=3.2*10-25/9.1*10-31

=0.351*106

=3.51*105m/s2

Q-4

We know that E=F/q

The equation tells us that if the amount of a charge of a test charge increase electric field will change but it is not like this.

 According to coulombs law if charges change means force will also change proportionally.

So in equation E=F/q

If in the above equation the denominator changes  by a factor 4 then the numerator will also change by the same factor 4 and hence electric field will remain same.

Q-5a

We know that gravitational force is a conservative force.  Work does not depend on the path taken. The P.E is shown by negative sign which means the energy spent in moving body from one point to another is stored in form of PE.  Similarly electric forces are also conservative force.

Therefore here

W=-∆U.=-10joule

 

(b) Is it true to say that electric potential energy and electric potential are the same?

Either yes or no explain in each case.

 

Answer. 5(b)

This not true to say both are same we can define them separately and then convince.

Electric potential energy is the energy stored in moving charges from a higher potential energy to a lower potential energy and it is in joule.

We can say electrical potential energy of a charge is the energy it has due to its position relative to other charges with which it interacts.

 

While electric potential is the electric potential energy per unit charge.

So Electric Potential=Electric potential energy/charge

Energy is measured in joule and charge in coulomb

If 1 joule energy is taken in moving 1 coulomb charge then

Electric Potential=1 joule/Coulomb

Which is 1 volt?

So both electric potential energy and electric potential are different not the same.

Q-3

It is true to say electric field lines are the same as electric field vector.

These electric field lines can be imagined by drawing field lines.   Electric field lines represent both the magnitude and direction electric field.

Electric field lines begin from positive and finish at negative. When the field lines are closed together, the field is dense and stronger. When the field line is apart the electric field is a weaker.

 

thanks

thanx :) :) :)

Muhammed Waseem Bangash Thanks for sharing 

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Attention Plzzzzz.

Q.1.b

K=9.0*109 Nm2/c2

q=12.5 C

r=11.5cm=11.5*10-2m here is mistake ...... 10 raise to power -4 not -2

Use the formula

E=K*q/r2

E=9.0*109 *12.5/(11.5*10-2) Nm2/C2 *C/m2

E=9*12.5/11.5*11.5*109*104N/C

E=0.85066*1013N/C

E=85.066*1011N/C   and here 10 raise to power 15 not 13 .......

Hope everybody got the solution....... :-)

Phy101 solved assigment 12/01/2013

chk kar lenaQuestion # 1
(a)What is the electric field at the location of the test charge shown below? And what is the
direction of electric field? Marks = 5
Physics (PHY101) VU of Pakistan
(b) What are the magnitude and direction of the electric field at point A shown below? Marks = 6
Question # 2
Find the force on the electron in figure shown below and also determine the acceleration of
electron? Mention the direction of force and acceleration as well.
Note that mass of electron is 9.1×10-31kg, and the value of charge on electron is (-1.6×10-19 C)
Marks = 7
Question # 3
Is it true to say that electric field lines are the same as electric field vectors. If yes give an
example if not explain it. Marks = 6
Question # 4
Suppose you increased the amount of the charge of a test charge, and the force exerted on it
changed. This means the field you were assessing must have changed. Marks = 5
Question # 5
(a) It takes 10.0 J of work by the right-hand wand to separate the charges as shown in figure.
What is the change in potential energy? Marks = 3
Physics (PHY101) VU of Pakistan
(b) Is it true to say that electric potential energy and electric potential are the same?
Either yes or no explain in each case. Marks = 3

Is it true to say that electric field lines are the same as electric field vectors. If yes give an
example if not explain it.
Answer:-
Yes It is true to say that electric field lines are the same as electric field vectors
A field line is a locus that is defined by a vector field and a starting location within the field.A vector field defines a direction at all points in space; a field line for that vector field may be constructed by tracing a path in the direction of the vector field. More precisely, the tangent line to the path at each point is required to be parallel to the vector field at that point.
A complete description of the geometery of all the field lines of a vector field is sufficient to completely specify the direction of the vector field everywhere. In order to also depict the magnitude, a selection of field lines is drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to the magnitude of the vector field at that point.



Q.1.a
E=F/q
use this formula to find Electric field.
Q.1.b
E=K.q/r2 (r square)
use this formula to find magnitude.
Q.3
Answer:-
Yes It is true to say that electric field lines are the same as electric field vectors
A field line is a locus that is defined by a vector field and a starting location within the field.A vector field defines a direction at all points in space; a field line for that vector field may be constructed by tracing a path in the direction of the vector field. More precisely, the tangent line to the path at each point is required to be parallel to the vector field at that point.
A complete description of the geometry of all the field lines of a vector field is sufficient to completely specify the direction of the vector field everywhere. In order to also depict the magnitude, a selection of field lines is drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to the magnitude of the vector field at that point.

Q.4
by the relation E=F/q it is very clear that if amount of charge or force is changed E will be changed.

Here is Q.5.b (wikipedia)
Electric potential energy, or electrostatic potential energy, is a potential energy (measured in joules) that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system.
The electric potential at a point is equal to the electric potential energy (measured in joules) of any charged particle at that location divided by the charge (measured in coulombs) of the particle. Since the charge of the test particle has been divided out, the electric potential is a "property" related only to the electric field itself and not the test particle. The electric potential can be calculated at a point in either a static (time-invariant) electric field or in a dynamic (varying with time) electric field at a specific time, and has the units of joules per coulomb, or volts.

Phy101 solved assigment 12/01/2013

(a):
F=1.2N
Q=2.5µc=2.5*10-6 Coulombs
Use the formula E=F/q
E=1.2/2.5*10-6
E=0.48*106
E=4.8*105N/c
E=480000N/C
(b): K=9.0*109 Nm2/c2
q=12.5 C
r=11.5cm=11.5*10-2m 
10 raise to power -4 not -2
formula
E=K*q/r2
E=9.0*109 *12.5/(11.5*10-2)2 Nm2/C2 *C/m2
E=9*12.5/11.5*11.5*109*104N/C
E=0.85066*1013N/C
E=85.066*1011N/C 

Q.NO:2 Find the force on the electron in figure show below and also determine the acceleration of electron? Mention the direction of force and acceleration as well.?
Note: the mass of electron is 〖9.31X10〗^(-31) kg and the value charge on electron is 〖–1.6X10〗^(-19) C

Ans: first we find the force 
f=m/a
F=E*Q here also 
So : m=f/a 
E=2.0*10-6N/C
Mass=m=9.1*10-31kg
Charge=q=-1.6*10-19C
By using formula
F=E*q
F=2.0*10-6*(-1.6*10-19)(N/C)© 
F=3.2*10-25N
For acceleration
Acceleration(a)=F/m
=3.2*10-25/9.1*10-31
=0.351*106
=3.51*105m/s2

q4

Field is force per unit charge .

When charge is increased ,force is also increases so as to keep the field constant.
Field is the same .

Cost 5 objects is 100$ .Hence the price is 20$ per object .

When the number is doubled the cost is doubled and it is 200$ .

The pirce remains the same as 20 $

Number is analogous to number of charges and the cost is analogous to the force and the field is analogous to the price.



Q.1.b
K=9.0*109 Nm2/c2
q=12.5 C
r=11.5cm=11.5*10-2m here is mistake ...... 10 raise to power -4 not -2
Use the formula
E=K*q/r2
E=9.0*109 *12.5/(11.5*10-2)Nm2/C2 *C/m2
E=9*12.5/11.5*11.5*109*104N/C
E=0.85066*1013N/C
E=85.066*1011N/C and here 10 raise to power 15 not 13 .......


SOLUTION:

Q.1.a

E=F/q
use this formula to find Electric field.

Q.1.b
E=K.q/r2 (r square)
use this formula to find magnitude.

Q.4


by the relation E=F/q it is very clear that if amount of charge or force is changed E will be changed.




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