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Expert-verified Found in: Page 669 ### College Physics (Urone)

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$$7.50 \times {10^2}\;V$$.

See the step by step solution

## 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 $$7.50 \times {10^2}\;V$$. ### Want to see more solutions like these? 