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Electricity and Magnetism VERY Basic Questions (no calculation ones just conceptual)

 
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May7-11, 11:46 AM   #1
 

Electricity and Magnetism VERY Basic Questions (no calculation ones just conceptual)

Yes, yes I know about the template, but these questions are so basic that I don't really know what I can show as work... Sometimes all I need for help is the relevant equation itself -.- So please don't delete or hate. Thanks!


2. http://tinyimage.net/images/50136173150204648365.png
The answer is A and A explain please?


4. http://tinyimage.net/images/85261815545593376175.png
umm the answer is A, and I have no idea what to use...

5. http://tinyimage.net/images/61071039590674115839.png
WHAT WHY IS ISN'T IT 2uJ? THIS DEFIES PHYSICS WAHHH joking lol

6. http://tinyimage.net/images/49749004950847017201.png
I'm guessing you use kirchoff's law for this...

7. http://tinyimage.net/images/00194558680517153372.png
so i get why 66 is B, but why is 67 B as well?? shouldn't the graph just be the formula shown in 66?

Any help would be appreciated
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May7-11, 11:58 AM   #2
 
Quote by compscier View Post
2. http://tinyimage.net/images/50136173150204648365.png
The answer is A and A explain please?
This really should not be answered, it is so basic. Just use coulomb's law

Quote by compscier View Post
3. How do you know which formula to use for the energy stored in the capacitor (you know there's 1/2QV and 1/2V^2C...) like for number 61 in http://tinyimage.net/images/82840419860896498375.png
It depends upon the ease of use. In 61 the capacitors are in series so q will be constant. Hence best would be 1/2Q2/C ---> You missed out on this formula
May7-11, 12:01 PM   #3
 
Quote by compscier View Post
1. a charged particle is projected parallel to a uniform to a magnetic field, the resulting path is a straight line parallel to the field... why? doesn't the right hand rule apply here
What does the right hand rule say?
May7-11, 12:14 PM   #4
 

Electricity and Magnetism VERY Basic Questions (no calculation ones just conceptual)

Quote by ashishsinghal View Post
This really should not be answered, it is so basic. Just use coulomb's law



It depends upon the ease of use. In 61 the capacitors are in series so q will be constant. Hence best would be 1/2Q2/C ---> You missed out on this formula
1. hmm is electric potential a scalar?

2. So wait, what's wrong here:
1/2*V^2*C=1/2*6*6*(1/(1/6+1/6+1/6))=1/2*36*2=18?
May7-11, 12:16 PM   #5
 
Quote by Abdul Quadeer View Post
What does the right hand rule say?

ahh is it because the angles between velocity of the particle and magnetic field is 0, thus eliminating the force?
May7-11, 12:24 PM   #6
 
Quote by compscier View Post
ahh is it because the angles between velocity of the particle and magnetic field is 0, thus eliminating the force?
Yes your are right.
It is convenient to use right hand rule to find out the direction of force acting on the particle when the magnetic field vector and velocity vector are perpendicular. You may answer this question by using the equation F=q(vxB)
May7-11, 12:32 PM   #7
 
Quote by compscier View Post
1. hmm is electric potential a scalar?

2. So wait, what's wrong here:
1/2*V^2*C=1/2*6*6*(1/(1/6+1/6+1/6))=1/2*36*2=18?
This will give you net energy. BTW 1/2*36*2=36
May7-11, 04:47 PM   #8
 
hmm help me on the other ones please?
May8-11, 01:15 PM   #9
 
Can't answer 5th.
For the 6th: just do it - nike
For the 7th: you need to tell why are you not getting B for 67
May8-11, 10:47 PM   #10
 
Quote by compscier View Post
5. http://tinyimage.net/images/61071039590674115839.png
WHAT WHY IS ISN'T IT 2uJ? THIS DEFIES PHYSICS WAHHH joking lol
2uJ is the work done by the Electric field. The magnetic field does no work.
May9-11, 12:17 PM   #11
 
Quote by Abdul Quadeer View Post
2uJ is the work done by the Electric field. The magnetic field does no work.
Yeah, that is the point. Since force is perpendicular to the velocity F.ds =0.
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