Question

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104 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)?

Answer 1

(a) 3000 V

For two parallel conducting plates, the potential difference between the plates is given by:

$\Delta V=Ed$

where

E is the magnitude of the electric field

d is the separation between the plates

Here we have:

$E=7.50\cdot 10^4 V/m$ is the electric field

d = 4.00 cm = 0.04 m is the distance between the plates

Substituting,

$\Delta V=(7.50\cdot 10^4 V/m)(0.04 m)=3000 V$

(b) 750 V

The potential difference between the two plates A and B is

$\Delta V = V_B - V_A = 3000 V$

Let's take plate A as the plate at 0 volts:

$V_A = 0 V$

The potential increases linearly going from plate A (0 V) to plate B (3000 V).

So, if the potential difference between A and B, separated by 4 cm, is 3000 V, then the potential difference between A and a point located at 1 cm from A is given by the proportion:

$3000 V : 4 cm = V(1 cm) : 1 cm$

and solving for V(1 cm) we find:

$V(1 cm)=\frac{(3000 V)(1 cm)}{4 cm}=750 V$

Answer 2

Potential difference is the difference in electrical potential. when the lowest potential is taken to be at zero volts then the potential 1.00 cm from that plate is 750 V.

What is the potential difference?

A potential difference is defined as the difference in the electrical potential between two points.

Given to us

Distance between the plate, d = 4 cm = 0.04 m

Electric field, E = 7.50×104 V/m .

A.) We know that for two parallel conducting plates, the potential difference between the plate is given as,

$\triangle V = Ed$

E = Magnitude of the electric field

d = distance between the plates

Substitute the values,

$E = 7.5 \times 10^4 \times 0.04\\\\E = 3000\rm\ V$

B.) We already know the potential difference between the two plates, therefore,

The potential difference between the two plates,

$\triangle V= V_A - V_B$

Since, one of the plates is having zero potential, therefore,

$\triangle V= V_A = 3000\rm\ V$

We know that the potential difference between the two plates is 300 which are 4 cm apart, therefore,

$\dfrac{\triangle V}{d} = \dfrac{V}{1}\\\\V = 750 \rm\ V$

Hence, when the lowest potential is taken to be at zero volts then the potential 1.00 cm from that plate is 750 V.

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