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Q15PE.

Expert-verifiedFound in: Page 669

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**The electric field strength between two parallel conducting plates separated by\(4.00\;cm\) is\(7.50 \times {10^4}\;V/m\). **

**(a) What is the potential difference between the plates? **

**(b) The plate with the lowest potential is taken to be at zero volts. What is the potential\(1.00\;cm\) from that plate (and\(3.00\;cm\) from the other)?**

The required solutions are

(a) Potential difference between two plates is\(3.00 \times {10^3}\;V\).

(b) The potential \(1.00\;cm\) from that plate is\(7.50 \times {10^2}\;V\).

- The distance between the plates is:

\(d = (4.00\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)\\ = 0.0400\;m end\)

- The electric field strength between the plates is:

\(E = 7.50 \times {10^4}\;V/m.\)

- The plate with the lower potential is taken to be at zero volts.

**The potential difference between two points separated by a distance **\(d\)** in a homogeneous electric field of magnitude **\(E\)** is **\(\Delta V = Ed{\rm{ }}......(1)\)**.**

(a)

Equation \((1)\) yields the potential difference between the two plates:

\(\Delta V = Ed\)

Filling in the values for \({\rm{E}}\)and\({\rm}\),

\(\Delta V = \left( {7.50 \times {{10}^4}\;V/m} \right)(0.0400\;m)\\ = 3.00 \times {10^3}\;V\)

Therefore, potential difference between two plates is \(3.00 \times {10^3}\;V\).

Substituting\(1.00\;cm\) for \(d\) into Equation \((1)\) yields the electric potential at a distance\(1.00\;cm\) from the plate with the lower potential (zero potential):

\(V = E(1.00\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)\\ = \left( {7.50 \times {{10}^4}\;V/m} \right)(0.0100\;m)\\ = 7.50 \times {10^2}\;V\ end\)

Therefore, the potential of the other plate is \(7.50 \times {10^2}\;V\).

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