The electric field strength between two parallel conducting plates sep

Pam Stokes

Pam Stokes

Answered question

2022-01-12

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104V. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate and 3.00 cm from the other?

Answer & Explanation

Beverly Smith

Beverly Smith

Beginner2022-01-13Added 42 answers

Step 1
a.
Identify the unknown: ​
The potential difference between the plates
Step 3
List the Knowns:
Electric field: E=7.5×104Vm
Distance between the two plates: d=4cm=0.04m
Step 4
Set Up the Problem: ​
Potential difference between two uniform metal plates:
ΔVAB=Ed
Step 5
Solve the Problem: ​
ΔVAB=7.5×104×0.04=3000V
Donald Cheek

Donald Cheek

Beginner2022-01-14Added 41 answers

Step 6
b.
Identify the unknown: ​
The potential 1 cm from the zero Volt plate
Step 8
Set Up the Problem: ​
ΔVAB=VAVB=VA0=VA=Ed
Step 9
Solve the Problem: ​
VA=7.5×104×0.01=750V

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?