# Electric Field and distance

WeiLoong

## 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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## Answers and Replies

Homework Helper
E=v/d
d=100/800
d=0.125m
That one is correct.
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.

WeiLoong
that mean ed = 800x0.40 only?

Homework Helper
Yes. Just, be careful with the proper sign of the potential.

WeiLoong
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

Homework Helper
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.
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.

WeiLoong
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 thats the only thing i can think about it.

Homework Helper
i guess it should be -320 since V is decreasing.
Yes, it should be -320, otherwise the E field vector cannot be uniform everywhere.
Q1= k(40uC)/12cm
No, that's still not correct. Please look up "Coulomb's law" either in your book or in the internet.

WeiLoong
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.