Question

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 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?

Final Answer

- $5.63\times 10^{3}\textrm{ V/m}$
- $563\textrm{ V}$

### Solution video

# OpenStax College Physics, Chapter 19, Problem 20 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Two parallel plates are separated by a distance of 10.0 centimeters and at some position—8.00 centimeters from the plate that we will take to be voltage zero or potential zero— there's a potential difference between this position and the reference plate of 450 volts and based on that information, we can figure out what the electric field is that's going to be pointing upwards between these plates and this electric field is constant everywhere so we can use that same electric field in part (b) to calculate what the voltage is between the plates themselves. So the electric field then is the voltage at this position—a distance

*l*from the zero plate— divided by the distance to get there and that's 450 volts divided by 8.00 times 10 to the minus 2 meters which is 5.63 times 10 to the 3 volts per meter. And then in part (b), the voltage between the plates themselves is gonna be that same electric field multiplied by the distance between the plates. So electric field times 10.0 times 10 to the minus 2 meters which is 563 volts.## Comments

WHY DID YOU USE 10 CM FOR PART B . I DIDNT GET IT

Hello, the question mentions that the plates are separated by 10cm, and Voltage is Electric field strength times separation between large conducting plates. This formula for voltage is only true with electric fields that don't change with distance, such as that found between two large conducting plates.

Hope that helps,

Shaun