Pam Stokes

2022-01-12

The electric field strength between two parallel conducting plates separated by 4.00 cm is $7.50\times {10}^{4}V$ . (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?

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:$E=7.5\times {10}^{4}\frac{V}{m}$

Distance between the two plates:$d=4cm=0.04m$

Step 4

Set Up the Problem:

Potential difference between two uniform metal plates:

$\mathrm{\Delta}{V}_{AB}=Ed$

Step 5

Solve the Problem:

$\mathrm{\Delta}{V}_{AB}=7.5\times {10}^{4}\times 0.04=3000V$

a.

Identify the unknown:

The potential difference between the plates

Step 3

List the Knowns:

Electric field:

Distance between the two plates:

Step 4

Set Up the Problem:

Potential difference between two uniform metal plates:

Step 5

Solve the Problem:

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:

$\mathrm{\Delta}{V}_{AB}={V}_{A}-{V}_{B}={V}_{A}-0={V}_{A}=Ed$

Step 9

Solve the Problem:

${V}_{A}=7.5\times {10}^{4}\times 0.01=750V$

b.

Identify the unknown:

The potential 1 cm from the zero Volt plate

Step 8

Set Up the Problem:

Step 9

Solve the Problem:

22+64

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