# Coulomb LawWatch

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#1 Attachment 802748802750

Question is in the attachment. Thank you for help .
0
4 months ago
#2
Field inside a conductor is 0
0<x<r, E = 0
x>r, E is inversely proportional to r^2
if field is E at x = r
should be E/4 at x = 2r

It is the negative gradient of the potential-distance graph
2
#3
(Original post by BobbJo)
Field inside a conductor is 0
0<x<r, E = 0
x>r, E is inversely proportional to r^2
if field is E at x = r
should be E/4 at x = 2r

It is the negative gradient of the potential-distance graph

Thank you for the reply . But my mark scheme said no field for 0 < x < r. Also it stated that at x = r, there is a discontinuity and a vertical line should be drawn.
0
4 months ago
#4
(Original post by esrever)
Thank you for the reply . But my mark scheme said no field for 0 < x < r. Also it stated that at x = r, there is a discontinuity and a vertical line should be drawn.
Yes that is what I said. Sorry for not being clear. Field is 0 for 0 < x < r. At x = r, field is E (vertical line) then it decreases according to the inverse square law.
0
#5
(Original post by BobbJo)
Yes that is what I said. Sorry for not being clear. Field is 0 for 0 < x < r. At x = r, field is E (vertical line) then it decreases according to the inverse square law.
Sorry I misunderstood the graph of x = r. But can you explain why is the graph a vertical line at x = r?
0
4 months ago
#6
(Original post by esrever)
Sorry I misunderstood the graph of x = r. But can you explain why the graph is a vertical line at x = r?
Inside the conductor, there is no field.
Just outside, there is a field. Therefore there is a discontinuity in the graph. It goes from 0 to the maximum field strength at x = r. It does not increase slowly from 0 to E, but it is a "sudden" increase.
Going further away, the field strength decreases according to inverse square law.

If you look at E as being the negative potential gradient of the potential graph, you can see that the gradient of the potential graph is 0 for 0 < x < r, then the gradient is maximum at x = r and decreases thereafter. Therefore there is a discontinuity. x=r is a "sharp point" in the potential graph. Thus a vertical line is drawn.
Last edited by BobbJo; 4 months ago
0
#7
(Original post by BobbJo)
Inside the conductor, there is no field.
Just outside, there is a field. Therefore there is a discontinuity in the graph. It goes from 0 to the maximum field strength at x = r. It does not increase smoothly from 0 to E, but it is a "sudden" increase.
Going further away, the field strength decreases according to inverse square law.

If you look at E as being the negative potential gradient, you can see that the gradient is 0 for 0 < x < r, then the gradient is maximum at x = r and decreases thereafter. Therefore there is a discontinuity. x=r is a "sharp point" in the potential graph. Thus a vertical line is drawn.
Thank you so much 0
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