Lamalam
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Stonebridge
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(Original post by Lamalam)
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Electric field strength is uniform.
Field strength is equated to (negative) potential gradient.
So the potential between the plates varies uniformly (straight line) with a gradient equal to the (negative) field strength.

The graph shows a negative gradient line.

In other words
the potential varies uniformly from one plate to the other
when the field is uniform.
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Lamalam
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(Original post by Stonebridge)
Electric field strength is uniform.
Field strength is equated to (negative) potential gradient.
So the potential between the plates varies uniformly (straight line) with a gradient equal to the (negative) field strength.

The graph shows a negative gradient line.

In other words
the potential varies uniformly from one plate to the other
when the field is uniform.
Why negative??
And does uniform mean same electric field?

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uberteknik
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(Original post by Lamalam)
Why negative??
And does uniform mean same electric field?

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The question says with reference to the +ve plate.

So at that +ve plate the field strength is maximum as shown on the y-axis.

The question also says the field varies uniformly between the plates. i.e. a straight line.

So the field strength at distance d from the +ve plate must be zero at the opposite plate. i.e. V=0 at distance d on the x-axis.

Join the two points and the gradient is -ve. i.e. reducing wrt to the +ve reference plate.
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Lamalam
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(Original post by uberteknik)
The question says with reference to the +ve plate.

So at that +ve plate the field strength is maximum as shown on the y-axis.

The question also says the field varies uniformly between the plates. i.e. a straight line.

So the field strength at distance d from the +ve plate must be zero at the opposite plate. i.e. V=0 at distance d on the x-axis.

Join the two points and the gradient is -ve. i.e. reducing wrt to the +ve reference plate.
Does uniform electric field mean Constant And does not change?

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Stonebridge
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(Original post by Lamalam)
Does uniform electric field mean Constant And does not change?

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Yes.
The value of E is equal everywhere in the field.
The force on a charge anywhere in the field is the same value and direction.
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Lamalam
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(Original post by Stonebridge)
Yes.
The value of E is equal everywhere in the field.
The force on a charge anywhere in the field is the same value and direction.
Why the force is constant ? One plate is positively charge one plate is negatively charged ? When a negatively charge moves towards the negatocely charge plate, isnt there an increasing repulsive force? I think I got confused about this and I did the question attached [email protected]@Name:  1402137618816.jpg
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Stonebridge
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The force on the charge, let's say it's positive, is made up of repulsion from the positive plate and attraction to the negative plate. So when you get nearer the positive plate, yes, the repulsion gets larger, but the attraction to the negative plate gets smaller. The result, in a uniform field, is that the total resultant force is constant.
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Zenarthra
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(Original post by Stonebridge)
The force on the charge, let's say it's positive, is made up of repulsion from the positive plate and attraction to the negative plate. So when you get nearer the positive plate, yes, the repulsion gets larger, but the attraction to the negative plate gets smaller. The result, in a uniform field, is that the total resultant force is constant.
(Original post by uberteknik)
The question says with reference to the +ve plate.

So at that +ve plate the field strength is maximum as shown on the y-axis.

The question also says the field varies uniformly between the plates. i.e. a straight line.

So the field strength at distance d from the +ve plate must be zero at the opposite plate. i.e. V=0 at distance d on the x-axis.

Join the two points and the gradient is -ve. i.e. reducing wrt to the +ve reference plate.
I jsut wanted to ask, if E=V/d will give the pd anywhere in a uniform electric field?

Thanks!
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Stonebridge
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(Original post by Zenarthra)
I jsut wanted to ask, if E=V/d will give the pd anywhere in a uniform electric field?

Thanks!
No it gives E, the field strength, anywhere in the field.
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(Original post by Stonebridge)
No it gives E, the field strength, anywhere in the field.
Is there a formula that gives pd anywhere is a uniform electric field?
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Lamalam
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(Original post by Stonebridge)
The force on the charge, let's say it's positive, is made up of repulsion from the positive plate and attraction to the negative plate. So when you get nearer the positive plate, yes, the repulsion gets larger, but the attraction to the negative plate gets smaller. The result, in a uniform field, is that the total resultant force is constant.
Thanks !

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Stonebridge
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(Original post by Zenarthra)
Is there a formula that gives pd anywhere is a uniform electric field?

You don't need a formula, surely.

If the field is uniform the pd varies uniformly from the one plate to the other.

If it's 100V on one plate and 0 V on the other, for example, and there was 100mm distance between the plates, the the pd would vary by 1 volt for every mm. So 20mm from the zero plate the potential would be 20V.
50mm (half way) it would be 50V. And so on. It's just simple proportion.
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(Original post by Stonebridge)
You don't need a formula, surely.

If the field is uniform the pd varies uniformly from the one plate to the other.

If it's 100V on one plate and 0 V on the other, for example, and there was 100mm distance between the plates, the the pd would vary by 1 volt for every mm. So 20mm from the zero plate the potential would be 20V.
50mm (half way) it would be 50V. And so on. It's just simple proportion.
Ok ok thanks, if it was negative then it be start - potential right?
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Stonebridge
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Just take the pd between the plates and divide it by the separation (V/d)
So if it was -20V on one plate and +20V on the other the pd is 40V
So if the separation was, say, 20mm there would be 40V/20mm = 2V for every mm.

So just take the distance from the one plate in mm and for every mm the pd changes by 2V

So every mm it goes 20,18,16,14, etc and then 0 at the centre and then -2, -4.-6 and so on to -20V
Just visualise it.
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