The electric field strength between two parallel conducting plates sep

Pam Stokes 2022-01-12 Answered
The electric field strength between two parallel conducting plates separated by 4.00 cm is \(\displaystyle{7.50}\times{10}^{{{4}}}{V}\). (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?

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Expert Answer

Beverly Smith
Answered 2022-01-13 Author has 2218 answers
Step 1
a.
Identify the unknown: ​
The potential difference between the plates
Step 3
List the Knowns:
Electric field: \(\displaystyle{E}={7.5}\times{10}^{{{4}}}\frac{{V}}{{m}}\)
Distance between the two plates: \(\displaystyle{d}={4}{c}{m}={0.04}{m}\)
Step 4
Set Up the Problem: ​
Potential difference between two uniform metal plates:
\(\displaystyle\Delta{V}_{{{A}{B}}}={E}{d}\)
Step 5
Solve the Problem: ​
\(\displaystyle\Delta{V}_{{{A}{B}}}={7.5}\times{10}^{{{4}}}\times{0.04}={3000}{V}\)
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Donald Cheek
Answered 2022-01-14 Author has 363 answers
Step 6
b.
Identify the unknown: ​
The potential 1 cm from the zero Volt plate
Step 8
Set Up the Problem: ​
\(\displaystyle\Delta{V}_{{{A}{B}}}={V}_{{{A}}}-{V}_{{{B}}}={V}_{{{A}}}-{0}={V}_{{{A}}}={E}{d}\)
Step 9
Solve the Problem: ​
\(\displaystyle{V}_{{{A}}}={7.5}\times{10}^{{{4}}}\times{0.01}={750}{V}\)
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