Calculating Electric Field and Distance for a Point Charge

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Homework Help Overview

The discussion revolves around calculating the electric field and potential difference associated with point charges, specifically focusing on the relationships between voltage, distance, and electric field strength.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for electric field and potential difference, questioning the correct distances and signs associated with the calculations. There are attempts to clarify the relationships between voltage and distance, as well as the application of Coulomb's law.

Discussion Status

Several participants are exploring different interpretations of the problem, particularly regarding the correct application of formulas and the implications of signs in their calculations. Some guidance has been offered regarding the need for careful consideration of physical situations and the proper use of equations.

Contextual Notes

There are indications of confusion regarding specific distances and the potential difference, as well as the need for further justification of answers based on the physical context of the problem.

WeiLoong
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Homework Statement


https://s.yimg.com/hd/answers/i/c7a04ad5e8d74b57a246295c815e183d_A.jpeg?a=answers&mr=0&x=1452518295&s=c54275383477f3e40a56996127dfee96

Homework Equations


Electric field, electrostatic

The Attempt at a Solution



E=v/d
d=100/800
d=0.125m

V=Ed
=800(0.125+0.4)
=420V
[/B]
 
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WeiLoong said:
E=v/d
d=100/800
d=0.125m
That one is correct.
WeiLoong said:
V=Ed
=800(0.125+0.4)
=420V
By potential at X, the question must refer to the potential difference between point X and the 0 V. So, 0.125+0.4 is not the correct value for the corresponding distance.
 
that mean ed = 800x0.40 only?
 
Yes. Just, be careful with the proper sign of the potential.
 
https://scontent-hkg3-1.xx.fbcdn.net/hphotos-xfp1/v/t1.0-9/12509720_10205411857847476_602511770052026952_n.jpg?oh=66b58b8024f86e46c7a9c860c4d89bcf&oe=56FB9605

neeed help about 15b.
W= kq/r + kq/r ?
no idea about it.hehe
 
WeiLoong said:
that mean ed = 800x0.40 only?
Just want to remind you, in case you are not yet aware of, that that answer is still missing something to be justified according to the physical situation given in the problem.
WeiLoong said:
neeed help about 15b.
W= kq/r + kq/r ?
no idea about it.hehe
Write down the formula of electric field due to each charge along the connecting line, define this line as ##x##. Denote ##x_1## and ##x_2## as the distances of a given point from ##Q_1## and ##Q_2##, respectively. The equation you have there is wrong.
 
Whats the missing part actually? i got 320 V. I am not sure it is -320 or +320V actually. but i guess it should be -320 since V is decreasing.

Q1= k(40uC)/12cm
Q2 =k(80uC)/12cm that's the only thing i can think about it.
 
WeiLoong said:
i guess it should be -320 since V is decreasing.
Yes, it should be -320, otherwise the E field vector cannot be uniform everywhere.
WeiLoong said:
Q1= k(40uC)/12cm
No, that's still not correct. Please look up "Coulomb's law" either in your book or in the internet.
 
Perhaps it should be kqq/r
can u explain this concept to me? i thought product of charges over distance is only used to determine the force, never thought u can be express as electric potential too.
 
  • #10
WeiLoong said:
Perhaps it should be kqq/r
There is no "perhaps" in this matter because the related equation which is sometimes called "Coulomb's law" had been agreed upon among the physicists long time ago. You will gain much more knowledge by searching/reading on your own about this basic concept of electric field by a point charge in electrostatics than you will if someone were to explain that to you.
 

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