# Electric field and voltage

## 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?

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?

Last edited:

Simon Bridge
Homework Helper
(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?

• 1 person
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
Homework Helper
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
Homework Helper
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