Q15PE. - College Physics (Urone)

Question

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)?

Step-by-Step Solution

Answer

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 given data

 

• 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.

Principle

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

Potential difference calculation

(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 \(3.00 \times {10^3}\;V\).

Part (b)

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 \(7.50 \times {10^2}\;V\).