Electric field strength between two disks?

[Corey Bacon]

Homework Statement

Two 3.0 cm diameter disks face each other, 1.9mm apart. They are charged to $$\pm 16nC$$.
(a). What is the field strength between the disks?
(b). A proton is shot from the negative disk toward the positive disk, what launch speed must the proton have to barely reach the positive disk?

I have attemped a solution for question (a), but MasteringPhysics says it is wrong.

Thanks heaps to anyone who can help
Corey

Homework Equations

The Attempt at a Solution

My attempt at question (a):
$$A = pi * r^2$$
$$r = 0.015$$
$$Q = 16*10^(-9)$$
$$E = Q/A*Epsilon_0 = Q/A*8.85*10^(-12) = 1.278*10^6 $$

 

[Qwertywerty]

I may be wrong , but you seem to have taken Q as charge on the entire disk , and not on just one face of it .

Hope this helps .

 

[haruspex]

Since the distance between the discs is very small compared with their diameter, you can treat it as two infinite parallel sheets. You need to involve the distance between them in the formula.

 

[Qwertywerty]

Since the distance between the discs is very small compared with their diameter, you can treat it as two infinite parallel sheets. You need to involve the distance between them in the formula.

The electric field between the two discs would be , approximately , σ / 2ε₀ . So why would you involve the distance between the plates ?

 

[Qwertywerty]

The electric field between the two discs would be , approximately , σ / 2ε0

Edit : σ / ε₀ .

 

[haruspex]

The electric field between the two discs would be , approximately , σ / 2ε₀ . So why would you involve the distance between the plates ?

Whoops, I was thinking in terms of voltage, not charge. Thanks for picking that up.
Corey, in your final step you seem to have divided by an extra 2. I would guess this is because you forgot there are two plates, one with positive charge and one negative. Their fields add.

 

[Qwertywerty]

Corey, in your final step you seem to have divided by an extra 2. I would guess this is because you forgot there are two plates, one with positive charge and one negative. Their fields add.

Actually , I believe he needs to divide further by two ( See post#2 ) .

 

[haruspex]

Actually , I believe he needs to divide further by two ( See post#2 ) .

It doesn't matter which side of a disc the charge is on, it will state generate the same field in the gap.

 

[Qwertywerty]

It doesn't matter which side of a disc the charge is on, it will state generate the same field in the gap.

The electric field in the gap will only be due to the charge present on discs , on the sides facing each other .

 

[haruspex]

The electric field in the gap will only be due to the charge present on discs , on the sides facing each other .

What property of the disc could block the contribution from the charge on the far side of the disc?
Also, bear in mind that if the discs conduct then the two lots of charges will attract each other and most will finish up on the inner surfaces anyway.

 

[Qwertywerty]

What property of the disc could block the contribution from the charge on the far side of the disc?

Hmm .. You're right . Thus I must correct my earler stance - the entire charge on the discs will lie on the sides facing each other .

And I've figured it out - the OP's just made a mistake in his calculations .