Calculating Electric Field and Voltage Between Parallel Conducting Plates

[tuggler]

Homework Statement

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts.

a. What is the electric field strength between them, if the potential 6.65 cm from the zero volt plate (and 3.35 cm from the other) is 633 V?

b. What is the voltage between the plates?

Homework Equations

V = Ed

The Attempt at a Solution

For a I got 9.5 keV which was from the formula: (633V)/0.0665m = 9.5eV E3.

For b I don't know what to do? The electric field is uniform so the electric field is just 9.5E3 and the distance will be 10cm/2 = 5 cm. But how can I get the correct answer?

 

[Simon Bridge]

(a) 9.5keV seems awfully low - check your units?
I don't see how you got that figure from your stated calcuation: 633/0.035=18086.

(b) you know the electric field between the plates and you know the separation, and you have an equation which relates these two values to the voltage. What's the problem?

Why are you dividing the 10cm separation between the plates by 2 to the "the distance"? What distance is that?
If you have 6V in 6cm, how many volts do you have in 10cm?
If there are 633 in 6.65cm, how many in 10cm?

 

[tuggler]

My mistake, its suppose to be 633/0.0665 = 9.5keV and it is the correct answer.

And I made a stupid mistake. I thought they said half way between the plates which is why I divided it by 2. Thank you very much! I know what to do now.

 

[Simon Bridge]

633V/0.0665m = 9.5keV

You appear to be out by a factor of about 10^23 - check units.
Apart from that - well done.

 

[ehild]

My mistake, its suppose to be 633/0.0665 = 9.5keV and it is the correct answer.

And I made a stupid mistake. I thought they said half way between the plates which is why I divided it by 2. Thank you very much! I know what to do now.

The first question is about the electric field strength, and its unit is V/m or kV/m, or N/C.
eV is unit of work or energy, about 1.6 E-19 joule.

ehild