# Electric field

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#2

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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(Original post by

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

when the field is uniform.

**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.

And does uniform mean same electric field?

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#4

(Original post by

Why negative??

And does uniform mean same electric field?

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**Lamalam**)Why negative??

And does uniform mean same electric field?

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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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(Original post by

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.

**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.

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#6

(Original post by

Does uniform electric field mean Constant And does not change?

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**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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(Original post by

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.

**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.

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#8

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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#9

(Original post by

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.

**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.

**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.

Thanks!

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#10

(Original post by

I jsut wanted to ask, if E=V/d will give the pd anywhere in a uniform electric field?

Thanks!

**Zenarthra**)I jsut wanted to ask, if E=V/d will give the pd anywhere in a uniform electric field?

Thanks!

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#11

(Original post by

No it gives E, the field strength, anywhere in the field.

**Stonebridge**)No it gives E, the field strength, anywhere in the field.

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**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.

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#13

(Original post by

Is there a formula that gives pd anywhere is a uniform electric field?

**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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#14

(Original post by

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.

**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.

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#15

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.

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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