What Is The Change In The Potential Energy Of The Dipole-field System In Part D?

Learning Objectives

Past the terminate of this section, you will be able to:

• Describe the human relationship betwixt voltage and electric field.
• Derive an expression for the electric potential and electric field.
• Calculate electrical field strength given distance and voltage.

Figure 1. The human relationship between V and E for parallel conducting plates is [latex]E=\frac{V}{d}\\[/latex]. (Note that ΔV = V _AB in magnitude. For a charge that is moved from plate A at college potential to plate B at lower potential, a minus sign needs to be included as follows: –ΔV = V _A – V _B = V _AB. See the text for details.)

In the previous section, we explored the relationship between voltage and energy. In this section, nosotros will explore the relationship between voltage and electrical field. For instance, a uniform electric field E is produced by placing a potential departure (or voltage) ΔV across ii parallel metal plates, labeled A and B. (See Figure 1.)

Examining this will tell u.s. what voltage is needed to produce a certain electric field force; it will besides reveal a more key relationship between electrical potential and electric field. From a physicist's bespeak of view, either ΔV or E tin can be used to draw any accuse distribution. ΔV is most closely tied to energy, whereas Due east is well-nigh closely related to strength. ΔV is a scalar quantity and has no direction, while Eastward is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electrical field strength, a scalar quantity, is represented by E beneath.) The relationship between ΔFive and E is revealed by calculating the work washed past the forcefulness in moving a charge from point A to point B.

Just, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electrical field as an interesting special example.

The work washed by the electric field in Figure one to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

W = −ΔPE = −qΔFive.

The potential deviation between points A and B is

−ΔV = −(V _B −V _A) =V _A −5 _B =V _AB.

Inbound this into the expression for work yieldsW =qV _AB.

Work is Due west =Fd cosθ; here cos θ = 1, since the path is parallel to the field, then W= Fd. Since F= qE, nosotros see that Westward= qEd. Substituting this expression for piece of work into the previous equation givesqEd= qV _AB.

The charge cancels, then the voltage betwixt points A and B is seen to be

[latex]\begin{cases}V_{\text{AB}}&=&Ed\\East&=&\frac{V_{\text{AB}}}{d}\end{cases}\\[/latex] (compatible Due east− field simply)

where d is the distance from A to B, or the distance between the plates in Figure 1. Annotation that the above equation implies the units for electrical field are volts per meter. Nosotros already know the units for electric field are newtons per coulomb; thus the following relation among units is valid: 1 N/C = 1 V/m.

Voltage between Points A and B

[latex]\brainstorm{cases}V_{\text{AB}}&=&Ed\\E&=&\frac{V_{\text{AB}}}{d}\end{cases}\\[/latex] (uniform E− field only)

where d is the distance from A to B, or the distance betwixt the plates.

Example 1. What Is the Highest Voltage Possible between 2 Plates?

Dry out air will support a maximum electric field strength of virtually three.0 × 10⁶ V/m. Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage betwixt two parallel conducting plates separated past two.5 cm of dry air?

Strategy

Nosotros are given the maximum electric field E betwixt the plates and the altitude d between them. The equation V _AB =Ed can thus exist used to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

Five _AB =Ed.

Entering the given values for E and d gives

V _AB = (3.0 × ten^half-dozen V/k)(0.025 m) 7.5 × 10⁴ 5 orV _AB = 75 kV.

(The reply is quoted to simply two digits, since the maximum field force is approximate.)

Word

One of the implications of this result is that it takes near 75 kV to brand a spark jump across a two.five cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between conductors, perhaps on a ability transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry out days.

Figure 2. A spark chamber is used to trace the paths of high-free energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between side by side plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)

Example ii. Field and Strength within an Electron Gun

1. An electron gun has parallel plates separated past 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between the plates?
2. What force would this field exert on a piece of plastic with a 0.500 μC charge that gets between the plates?

Strategy

Since the voltage and plate separation are given, the electric field forcefulness can be calculated directly from the expression [latex]E=\frac{V_{\text{AB}}}{d}\\[/latex]. Once the electrical field strength is known, the force on a accuse is institute using F = q East. Since the electric field is in merely one direction, we can write this equation in terms of the magnitudes, F =qE.

Solution for Part 1

The expression for the magnitude of the electric field between two compatible metallic plates is

[latex]E=\frac{V_{\text{AB}}}{d}\\[/latex].

Since the electron is a single charge and is given 25.0 keV of energy, the potential deviation must exist 25.0 kV. Inbound this value for 5 _AB and the plate separation of 0.0400 one thousand, we obtain

[latex]E=\frac{25.0\text{ kV}}{0.0400\text{ m}}=6.25\times10^5\text{ 5/m}\\[/latex]

Solution for Part ii

The magnitude of the force on a charge in an electric field is obtained from the equationF=qE.

Substituting known values gives

F = (0.500 × 10⁻⁶ C)(six.25 × 10⁵ V/k) = 0.313 N.

Discussion

Note that the units are newtons, since 1 V/m = 1 N/C. The force on the accuse is the same no matter where the charge is located between the plates. This is because the electric field is uniform between the plates.

In more than general situations, regardless of whether the electric field is compatible, information technology points in the direction of decreasing potential, considering the force on a positive charge is in the direction of Eastward and also in the direction of lower potential V. Furthermore, the magnitude of E equals the rate of decrease of V with distance. The faster V decreases over distance, the greater the electric field. In equation grade, the general relationship between voltage and electric field is

[latex]E=-\frac{\Delta{Five}}{\Delta{s}}\\[/latex],

where Δsouthward is the altitude over which the change in potential, ΔV, takes identify. The minus sign tells us that E points in the direction of decreasing potential. The electrical field is said to be the gradient (as in grade or slope) of the electric potential.

Relationship between Voltage and Electric Field

In equation form, the general human relationship between voltage and electric field is

[latex]\displaystyle{Eastward}=-\frac{\Delta{V}}{\Delta{s}}\\[/latex],

where Δs is the distance over which the change in potential, ΔFive, takes identify. The minus sign tells us that Due east points in the direction of decreasing potential. The electric field is said to exist the gradient (equally in grade or gradient) of the electric potential.

For continually changing potentials, Δ5 and Δs become infinitesimals and differential calculus must be employed to determine the electric field.

Department Summary

• The voltage between points A and B is

[latex]\begin{cases}V_{\text{AB}}&=&Ed\\E&=&\frac{V_{\text{AB}}}{d}\end{cases}\\[/latex] (uniform E− field only)

where d is the distance from A to B, or the distance betwixt the plates.

• In equation form, the general relationship between voltage and electric field is

[latex]E=-\frac{\Delta{V}}{\Delta{due south}}\\[/latex],

where Δs is the distance over which the change in potential, ΔV, takes identify. The minus sign tells u.s.a. that Due east points in the direction of decreasing potential.) The electrical field is said to exist the slope (every bit in grade or slope) of the electrical potential.

Conceptual Questions

1. Discuss how potential difference and electrical field force are related. Give an example.
2. What is the strength of the electric field in a region where the electric potential is constant?
3. Will a negative accuse, initially at residue, move toward higher or lower potential? Explain why.

Problems & Exercises

1. Testify that units of Five/m and N/C for electric field strength are indeed equivalent.
2. What is the strength of the electric field between two parallel conducting plates separated past one.00 cm and having a potential difference (voltage) betwixt them of 1.50 × ten⁴ V?
3. The electric field forcefulness betwixt two parallel conducting plates separated by four.00 cm is 7.fifty × 10⁴ V/k. (a) What is the potential divergence between the plates? (b) The plate with the lowest potential is taken to be at cypher volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?
4. How far autonomously are two conducting plates that have an electric field strength of 4.50× 10³ V/thou between them, if their potential deviation is 15.0 kV?
5. (a) Volition the electric field strength between ii parallel conducting plates exceed the breakup strength for air (3.0 × x⁶ Five/m) if the plates are separated by two.00 mm and a potential difference of 5.0 × 10^three V is applied? (b) How shut together can the plates be with this applied voltage?
6. The voltage across a membrane forming a cell wall is eighty.0 mV and the membrane is 9.00 nm thick. What is the electric field forcefulness? (The value is surprisingly big, only right. Membranes are discussed in Capacitors and Dielectrics and Nerve Conduction—Electrocardiograms.) Y'all may assume a compatible electric field.
7. Membrane walls of living cells have surprisingly large electric fields beyond them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms.) What is the voltage across an 8.00 nm–thick membrane if the electrical field strength beyond it is five.l MV/m? You may presume a compatible electric field.
8. Ii parallel conducting plates are separated past 10.0 cm, and one of them is taken to be at nil volts. (a) What is the electrical field force betwixt them, if the potential eight.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?
9. Observe the maximum potential difference between two parallel conducting plates separated past 0.500 cm of air, given the maximum sustainable electric field force in air to exist three.0 × 10⁶ V/chiliad.
10. A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field betwixt ii parallel conducting plates separated by 2.00 cm. What is the electric field strength betwixt the plates?
11. An electron is to exist accelerated in a uniform electric field having a strength of two.00 × 10⁶ V/chiliad. (a) What free energy in keV is given to the electron if it is accelerated through 0.400 chiliad? (b) Over what distance would it accept to be accelerated to increase its energy by 50.0 GeV?

Glossary

scalar: physical quantity with magnitude but no management

vector: physical quantity with both magnitude and direction

Selected Solutions to Problems & Exercises

3. (a) three.00 kV; (b) 750 V

5. (a) No. The electric field strength between the plates is ii.v × 10⁶ V/m, which is lower than the breakdown strength for air (iii.0 × 10^six V/thousand}); (b) i.7 mm

7. 44.0 mV

nine. xv kV

11. (a) 800 KeV; (b) 25.0 km

Source: https://courses.lumenlearning.com/physics/chapter/19-2-electric-potential-in-a-uniform-electric-field/

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