PHYSICS 1251 Lecture 7: Lecture Note Jan 22, 2019

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mauvegoldfish544

26 Jan 2019

School

Ohio State University

Department

Physics

Course

PHYSICS 1251

Professor

Tin-Lun Ho

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PHYSICS 1251 Lecture Notes - Lecture 6: Field Line, Electric Field, Gaussian Surface

PHYSICS 1251 Lecture 7: Lecture Note Jan 22, 2019

PHYSICS 1251 Lecture 8: Electric Fields Quiz Practice

Document Summary

The electric field is equal electric force rector divided by the magnitude of. Basically the from the position charge the charge power of the vector e to the go q q. 1 p e r sourcecharge ke q q f. Recalli ke 8. 99109 f unitrector in the direction of. Notice that the test charge q was eliminated. Pike field is only dependent on and the position from the the source charge source charge. The number of field lines is proprtimal to the charges. Notice how the large charge has twice the amount of pines. The particle is floating midair calculate the electric field on q. Masso4. 02g q 17 l fg. fo. oouoakglk. my f feq 9. 3934,442319. 776 epgwn. warpe. Theforceis positive but it is dividedby a negative charge considerthe three parts along the circle arc. E 3rd ez 3nc find the electric field at point p. Note thefields are broken up in their x and y direct i ii w a.

Related Questions

I am studying for a physics exam next Monday and I need tounderstand this concept a hell of lot better than I do. Hopefullysomeone on here can dumb it down for me.

My textbook says the following:

1. If the probe charge q is positive, the electric fieldvector points in the same direction as the force on the charge; ifnegative, the electric field vector points opposite theforce.

What is this really saying? I don't understand.

2. For the equation, Electric field = K|q|/r^2 , theelectric field is going away from q if q>0, and the electricfield is going toward q if q<0.

I don't understand this either. Is the electric field going awayfrom q because the electric field is negative and q is positive,thus q is attracted to the field? If q is negative then the fieldis going to be going toward q because it's being repelled from thenegative field?

I just want tounderstand the two bolded concepts above and how they relate toeach other, if they in fact do. My job does a horrible job atexplaining it.

Thanks.

This has been bothering me for a long time and I cant find anadequate explanation..

Lets say you have an infinite conducting plate with finitethickness D and it is given charge Q

the charge will be equally split into two and concentrated only onthe two surfaces.

Each surface will get Q/2

and the magnitude of electric field outside the plate isE=Q/(2A*epsilon naught) (E for a conducting plate is surface chargedensity/epsilon naught as opposed to surface chargedensity/(2epsilon naught)

Then, if you have two nonconducting plates separated by distance Dand each plate gets Q/2 which is uniformly distributed ..
then outside the space between the two plates, E = 2*(Q/4A*epsilonnaught)

But why cant we use superposition principle in the case of oneconducting plate with Q and say you get E=Q/(A epsilon naught)i.e.Q/(2A epsilon naught) from each surface...

Is it because we are looking at the mid section of the conductor assome kind of Faraday cage? (since the plate is essentially infinityand blocks the electric field from the other side?)

Also, lets look at two parallel plate capacitor..

One plate gets +Q/2 and the other gets -Q/2

I know the formula says E= charge density/epsilon naught.. so E hasto be Q/(2A epsilon naught) However, it seems rather inconsistentin the case of the parallel plate capacitor...

We know all the charge is going to be concentrated on the innersurface
So shouldnt you get E = Q/(2A epsilon naught) solely from one plateand since you have TWO plates!!!, shoudlnt you multiply Q/(2Aepsilon naught) by 2?

two point charges, q1=-2 microC and q2= -2microC are located onthe y axis. The charge at q1 is located at y=+1m and charge q2 aty=-1m.

1. Determine the electric field at point P on the x-axis atx=0.5m. specify direction

2. calculated the electric force on a point charge particle withq3=3 microC placed at P. specify direction

3. if you place q3 in item (b) at the origin of the coordinatesystem O and release it, the charge remains at that position, forthe net force on it is zero. You now displace q3 by a smalldistance along the x-axis and release it again. does charge q3: (i)remain at the new position (ii) move further away from origin (iii)return to origin. justify your answer

4. a student replaces charge q'2 by a unknown charge q2. placingnow charge q3=3microC at y=0.5 on the y axis, he notes that q3remains at rest. find the value of q'2

PHYSICS 1251 Lecture 7: Lecture Note Jan 22, 2019

PHYSICS 1251 Lecture Notes - Lecture 6: Field Line, Electric Field, Gaussian Surface

PHYSICS 1251 Lecture 7: Lecture Note Jan 22, 2019

PHYSICS 1251 Lecture 8: Electric Fields Quiz Practice

Document Summary

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Related textbook solutions

Physics: Principles with Applications

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