# Electric field strength between two disks?

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1. Aug 10, 2015

### Corey Bacon

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

Last edited: Aug 10, 2015
2. Aug 10, 2015

### 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 .

3. Aug 10, 2015

### 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.

4. Aug 10, 2015

### Qwertywerty

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

5. Aug 11, 2015

### Qwertywerty

Edit : σ / ε0 .

6. Aug 11, 2015

### haruspex

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.

7. Aug 11, 2015

### Qwertywerty

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

8. Aug 11, 2015

### haruspex

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

9. Aug 11, 2015

### Qwertywerty

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

10. Aug 11, 2015

### haruspex

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.

Last edited: Aug 11, 2015
11. Aug 11, 2015

### Qwertywerty

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 .