# Physics electrostatics

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Can anyone tell me why the answer is C instead of D? Thanks!

and one more question: my teacher told me that when a charge of fixed charge density is put between two charges paralla plates (of different charge), THe Electric field is independent of the area of plates and the separation between plates.

but there is also an equation of E=V/d.

aren't they contradict with each other? can someone explain it to me???

thank you!!!!!!

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#2

(Original post by

Can anyone tell me why the answer is C instead of D? Thanks!

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**Lamalam**)Can anyone tell me why the answer is C instead of D? Thanks!

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Thanks!

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(Original post by

Thank you, but what is the 1.6 in the equation?

**Zenarthra**)Thank you, but what is the 1.6 in the equation?

btw do you know the answer to my questions??

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#8

(Original post by

I sub 1.6 as the g because the question states that the loop is situated on moon.

btw do you know the answer to my questions??

**Lamalam**)I sub 1.6 as the g because the question states that the loop is situated on moon.

btw do you know the answer to my questions??

I think, it is 2 points because you place a positive test charge on the line.

See here:

The one in red, the forces act in opposite direction and the black one they act in same direction.

But im not sure sorry.

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#9

(Original post by

Can anyone tell me why the answer is C instead of D? Thanks!

and one more question: my teacher told me that when a charge of fixed charge density is put between two charges paralla plates (of different charge), THe Electric field is independent of the area of plates and the separation between plates.

but there is also an equation of E=V/d.

aren't they contradict with each other? can someone explain it to me???

thank you!!!!!!

Posted from TSR Mobile

**Lamalam**)Can anyone tell me why the answer is C instead of D? Thanks!

and one more question: my teacher told me that when a charge of fixed charge density is put between two charges paralla plates (of different charge), THe Electric field is independent of the area of plates and the separation between plates.

but there is also an equation of E=V/d.

aren't they contradict with each other? can someone explain it to me???

thank you!!!!!!

Posted from TSR Mobile

You may have misheard your teacher.

The capacitance of a parallel plate capacitor is given by

C = ϵA/d to a good approximation, so that depends on the area of the plates.

The field between them is given by E = V/d as you say.

Those are the facts.

So the field shouldn't depend on the area of the plates so long as you do not go too near the edges of the plates where the field lines start to bend.

The field strength E does depend on the separation so long as V is constant. You need to check with your teacher exactly what he/she was actually referring to.

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(Original post by

You may have misheard your teacher.

The capacitance of a parallel plate capacitor is given by

C = ϵA/d to a good approximation, so that depends on the area of the plates.

The field between them is given by E = V/d as you say.

Those are the facts.

So the field shouldn't depend on the area of the plates so long as you do not go too near the edges of the plates where the field lines start to bend.

The field strength E does depend on the separation so long as V is constant. You need to check with your teacher exactly what he/she was actually referring to.

**Stonebridge**)You may have misheard your teacher.

The capacitance of a parallel plate capacitor is given by

C = ϵA/d to a good approximation, so that depends on the area of the plates.

The field between them is given by E = V/d as you say.

Those are the facts.

So the field shouldn't depend on the area of the plates so long as you do not go too near the edges of the plates where the field lines start to bend.

The field strength E does depend on the separation so long as V is constant. You need to check with your teacher exactly what he/she was actually referring to.

This is stated in my note

What is he trying to express?

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(Original post by

Thank you for that!

I think, it is 2 points because you place a positive test charge on the line.

See here:

The one in red, the forces act in opposite direction and the black one they act in same direction.

But im not sure sorry.

**Zenarthra**)Thank you for that!

I think, it is 2 points because you place a positive test charge on the line.

See here:

The one in red, the forces act in opposite direction and the black one they act in same direction.

But im not sure sorry.

what are the first two sentences about?

and the third line: what is E referring to ?? the total electric field??

thankksss!!! I start to understnad a bit now!!

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#12

(Original post by

This is stated in my note

What is he trying to express?

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**Lamalam**)This is stated in my note

What is he trying to express?

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So you halve the distance d but the result is you also halve the pd.

So E doesn't change.

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#13

(Original post by

what are the first two sentences about?

and the third line: what is E referring to ?? the total electric field??

thankksss!!! I start to understnad a bit now!!

**Lamalam**)what are the first two sentences about?

and the third line: what is E referring to ?? the total electric field??

thankksss!!! I start to understnad a bit now!!

So maybe there is a point where Fa force due to point charge A on the positive test charge and Fb the force due to point charge B on the positive test charge are equal in magnitude but not direction.

E at a point = Resultant F on positive test charge at that point / charge of positive test charge q

As you can see positive test charge is constant q

So if forces at a point due to each point charge are equal the Electric field due to each point charge at that point are equal.

Force direction is dependent on the charges.

Electric field is vector quantity so at that point you would have E=0 if you add the electric fields together.

For the Black writing, if the positive test charge is between the 2 point charges, the positive test charge will experience a force due to point charge A and B.

The direction of the force is in the same direction.

E at a point = Resultant F on positive test charge at that point / charge of positive test charge q.

They are in the same direction so the resultant electric field at some point between them can never be 0.

hence the electric field at some point between them can not be = 0

But Fa could = Fb or just 2Fa or 2Fb, so that the resultant force = Fa + Fb.

Where the electric field at any point between the point charges = Fa + Fb / q

because q is constant.

! No problem!

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(Original post by

Yes if you look at the red writing, you can see that the force acting on the test positive charge will be opposite in direction to each other.

So maybe there is a point where Fa force due to point charge A on the positive test charge and Fb the force due to point charge B on the positive test charge are equal in magnitude but not direction.

E at a point = Resultant F on positive test charge at that point / charge of positive test charge q

As you can see positive test charge is constant q

So if forces at a point due to each point charge are equal the Electric field due to each point charge at that point are equal.

Force direction is dependent on the charges.

Electric field is vector quantity so at that point you would have E=0 if you add the electric fields together.

For the Black writing, if the positive test charge is between the 2 point charges, the positive test charge will experience a force due to point charge A and B.

The direction of the force is in the same direction.

E at a point = Resultant F on positive test charge at that point / charge of positive test charge q.

They are in the same direction so the resultant electric field at some point between them can never be 0.

hence the electric field at some point between them can not be = 0

But Fa could = Fb or just 2Fa or 2Fb, so that the resultant force = Fa + Fb.

Where the electric field at any point between the point charges = Fa + Fb / q

because q is constant.

! No problem!

**Zenarthra**)Yes if you look at the red writing, you can see that the force acting on the test positive charge will be opposite in direction to each other.

So maybe there is a point where Fa force due to point charge A on the positive test charge and Fb the force due to point charge B on the positive test charge are equal in magnitude but not direction.

E at a point = Resultant F on positive test charge at that point / charge of positive test charge q

As you can see positive test charge is constant q

So if forces at a point due to each point charge are equal the Electric field due to each point charge at that point are equal.

Force direction is dependent on the charges.

Electric field is vector quantity so at that point you would have E=0 if you add the electric fields together.

For the Black writing, if the positive test charge is between the 2 point charges, the positive test charge will experience a force due to point charge A and B.

The direction of the force is in the same direction.

E at a point = Resultant F on positive test charge at that point / charge of positive test charge q.

They are in the same direction so the resultant electric field at some point between them can never be 0.

hence the electric field at some point between them can not be = 0

But Fa could = Fb or just 2Fa or 2Fb, so that the resultant force = Fa + Fb.

Where the electric field at any point between the point charges = Fa + Fb / q

because q is constant.

! No problem!

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#15

Wey, ley doe how lek ah!

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(Original post by

Well the text says a constant charge density on the plates. Not a constant charge between the plates. Yes E depends on d if V is kept constant, as I said. But reducing d for example will increase the capacitance. Halving d doubles the capacitance according to the formula C = ϵA/d. If you double the capacitance with constant Q then you halve the pd from Q = CV

So you halve the distance d but the result is you also halve the pd.

So E doesn't change.

**Stonebridge**)Well the text says a constant charge density on the plates. Not a constant charge between the plates. Yes E depends on d if V is kept constant, as I said. But reducing d for example will increase the capacitance. Halving d doubles the capacitance according to the formula C = ϵA/d. If you double the capacitance with constant Q then you halve the pd from Q = CV

So you halve the distance d but the result is you also halve the pd.

So E doesn't change.

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#17

(Original post by

I am still a bit confused can you explain more ? thank you

**Lamalam**)I am still a bit confused can you explain more ? thank you

If you halve d, the distance between the plates, and the result is that you also halve V, the pd between the plates, then the field strength E doesn't change because E = V/d

All you've done is get

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(Original post by

I'm not sure what you are stuck on.

If you halve d, the distance between the plates, and the result is that you also halve V, the pd between the plates, then the field strength E doesn't change because E = V/d

All you've done is get

**Stonebridge**)I'm not sure what you are stuck on.

If you halve d, the distance between the plates, and the result is that you also halve V, the pd between the plates, then the field strength E doesn't change because E = V/d

All you've done is get

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#19

(Original post by

I thought there is only electric field strength between the the parallal plates, there is also electric field strength on the plate?!

**Lamalam**)I thought there is only electric field strength between the the parallal plates, there is also electric field strength on the plate?!

Are you sure you understand what electric field strength is?

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(Original post by

Are you sure you understand what electric field strength is?

**Stonebridge**)Are you sure you understand what electric field strength is?

A region of space in which a charged object experience a force?

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