You are watching: Suppose you have a uniform electric field

Figure 1. The relationship between *V* and also *E* for parallel conducting key is *V* = *V*AB in magnitude. Because that a fee that is moved from plate A at higher potential come plate B at lower potential, a minus sign requirements to be had as follows: –Δ*V* = *V*A – *V*B = *V*AB. View the text for details.)

In the previous section, we explored the relationship between voltage and energy. In this section, us will explore the relationship between voltage and also electric field. Because that example, a uniform electrical field **E** is produced by put a potential difference (or voltage) Δ*V* throughout two parallel metal plates, labeled A and B. (See number 1.)

Examining this will tell us what voltage is required to produce a certain electric field strength; the will additionally reveal a more an essential relationship in between electric potential and also electric field. From a physicist’s point of view, either Δ*V* or **E** deserve to be used to describe any charge distribution. Δ*V* is most carefully tied come energy, whereas **E** is most carefully related to force. Δ*V* is a *scalar* quantity and also has no direction, while **E** is a *vector* quantity, having both magnitude and also direction. (Note that the magnitude of the electric field strength, a scalar quantity, is stood for by *E* below.) The relationship in between Δ*V* and also **E** is revealed by calculating the work done by the pressure in relocating a charge from suggest A to allude B.

But, as noted in electric Potential Energy: Potential Difference, this is facility for arbitrary charge distributions, request calculus. We because of this look at a uniform electric field together an amazing special case.

The occupational done by the electric field in number 1 to relocate a positive charge *q * native A, the positive plate, higher potential, to B, the negative plate, reduced potential, is

*W* = −ΔPE = −*q*Δ*V*.

The potential difference in between points A and also B is

−Δ*V* = −(*V*B − *V*A) = *V*A − *V*B = *V*AB.

Entering this right into the expression for occupational yields *W* = *qV*AB.

Work is *W* = *Fd* cos *θ*; here cos * θ* = 1, because the course is parallel to the field, and so *W *= Fd. Due to the fact that *F *= qE, we watch that *W *= *qEd*. Substituting this expression because that work right into the vault equation gives *qEd *= *qV*AB.

The fee cancels, and so the voltage in between points A and also B is seen to be

*E *− ar only)

where *d* is the distance from A come B, or the distance between the bowl in figure 1. Keep in mind that the above equation means the devices for electrical field room volts per meter. We already know the units for electric field room newtons per coulomb; for this reason the following relation amongst units is valid: 1 N/C = 1 V/m.

### Voltage between Points A and B

*E *− field only)

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

### Example 1. What Is the greatest Voltage feasible between two Plates?

Dry waiting will support a maximum electrical field strength of around 3.0 × 106 V/m. Above that value, the ar creates sufficient ionization in the waiting to do the waiting a conductor. This permits a discharge or spark that reduces the field. What, then, is the best voltage in between two parallel conducting key separated through 2.5 cm of dried air?

StrategyWe are offered the maximum electrical field *E* in between the plates and the street *d* in between them. The equation *V*AB = *Ed* have the right to thus be supplied to calculate the preferably voltage.

The potential difference or voltage in between the key is

*V*AB = *Ed*.

Entering the given values because that *E* and also *d* gives

*V*AB = (3.0 × 106 V/m)(0.025 m) 7.5 × 104 V or *V*AB = 75 kV.

(The price is quoted to just two digits, due to the fact that the maximum ar strength is approximate.)

DiscussionOne the the effects of this an outcome is that it takes around 75 kV to make a spark jump throughout a 2.5 centimeter (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that have the right to exist between conductors, probably on a power transmission line. A smaller voltage will reason a spark if there space points on the surface, because points develop greater areas than smooth surfaces. Humid waiting breaks under at a lower field strength, definition that a smaller voltage will make a spark jump with humid air. The largest voltages have the right to be developed up, say v static electricity, on dried days.

Figure 2. A spark room is used to trace the routes of high-energy particles. Ionization developed by the particles as they pass v the gas in between the plates permits a spark come jump. The sparks are perpendicular to the plates, following electrical field lines between them. The potential distinction between nearby plates is not high sufficient to cause sparks without the ionization produced by particles from accelerator experiment (or cosmic rays). (credit: Daderot, Wikimedia Commons)

### Example 2. Field and Force within an Electron Gun

An electron gun has parallel bowl separated through 4.00 cm and gives electron 25.0 keV of energy. What is the electric field strength between the plates?What force would this ar exert ~ above a item of plastic v a 0.500 μC fee that gets in between the plates?StrategySince the voltage and plate separation space given, the electric field strength deserve to be calculated straight from the expression **F*** = q***E**. Due to the fact that the electrical field is in just one direction, we can write this equation in regards to the magnitudes, *F* = *qE*.

The expression because that the magnitude of the electric field in between two uniform steel plates is

Since the electron is a single charge and also is provided 25.0 keV that energy, the potential distinction must it is in 25.0 kV. Beginning this value for *V*AB and the bowl separation of 0.0400 m, we obtain

The magnitude of the pressure on a fee in an electrical field is acquired from the equation *F *= *qE*.

Substituting recognized values gives

*F* = (0.500 × 10−6 C)(6.25 × 105 V/m) = 0.313 N.

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

In much more general situations, regardless of even if it is the electric field is uniform, it points in the direction of to decrease potential, due to the fact that the pressure on a hopeful charge is in the direction the **E** and likewise in the direction of reduced potential *V*. Furthermore, the size of **E** amounts to the price of diminish of *V* v distance. The quicker *V* decreases end distance, the greater the electric field. In equation form, the general relationship in between voltage and electric ar is

where Δ*s* is the street over i beg your pardon the readjust in potential, Δ*V*, take away place. The minus sign tells united state that **E** clues in the direction of decreasing potential. The electric field is stated to be the *gradient* (as in grade or slope) of the electric potential.

### Relationship between Voltage and Electric Field

In equation form, the basic relationship between voltage and also electric ar is

where Δ*s* is the distance over i m sorry the change in potential, Δ*V*, takes place. The minus authorize tells us that **E** clues in the direction of diminish potential. The electrical field is stated to be the *gradient* (as in grade or slope) the the electric potential.

For continually transforming potentials, Δ*V* and Δ*s* come to be infinitesimals and also differential calculus should be to work to determine the electric field.

## Section Summary

The voltage between points A and also B is*E *− field only)

where *d* is the distance from A come B, or the distance in between the plates.

where Δ*s* is the distance over i beg your pardon the change in potential, Δ*V*, bring away place. The minus sign tells united state that **E** points in the direction of to decrease potential.) The electric field is stated to be the gradient (as in great or slope) that the electric potential.

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### Conceptual Questions

Discuss just how potential difference and also electric field strength are related. Offer an example.What is the toughness of the electric field in a region where the electrical potential is constant?Will a an adverse charge, originally at rest, relocate toward higher or reduced potential? explain why.### Problems & Exercises

Show that units of V/m and also N/C for electric field strength are undoubtedly equivalent.What is the strength of the electric field in between two parallel conducting key separated through 1.00 cm and having a potential difference (voltage) between them that 1.50 × 104 V?The electric field strength between two parallel conducting bowl separated by 4.00 centimeter is 7.50 × 104 V/m. (a) What is the potential difference in between the plates? (b) The plate v the shortest potential is taken to be in ~ zero volts. What is the potential 1.00 centimeter from that plate (and 3.00 cm from the other)?How far apart space two conducting bowl that have an electrical field strength of 4.50× 103 V/m in between them, if your potential difference is 15.0 kV?(a) will the electric field strength in between two parallel conducting key exceed the break down strength because that air (3.0 × 106 V/m) if the plates space separated by 2.00 mm and also a potential distinction of 5.0 × 103 V is applied? (b) exactly how close together have the right to the plates be v this applied voltage?Two parallel conducting plates space separated by 10.0 cm, and also one of them is taken to be at zero volts. (a) What is the electrical field strength in between them, if the potential 8.00 cm from the zero volt key (and 2.00 centimeter from the other) is 450 V? (b) What is the voltage between the plates?Find the preferably potential difference between two parallel conducting plates separated by 0.500 cm of air, offered the maximum sustainable electric field strength in waiting to it is in 3.0 × 106 V/m.A doubly charged ion is increased to an power of 32.0 keV by the electrical field in between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?An electron is to be accelerated in a uniform electrical field having actually a toughness of 2.00 × 106 V/m. (a) What power in keV is offered to the electron if it is sped up through 0.400 m? (b) end what distance would certainly it need to be sped up to rise its energy by 50.0 GeV?## Glossary

**scalar:** physical quantity with magnitude yet no direction

**vector:** physical quantity with both magnitude and also direction

### Selected remedies to Problems & Exercises

3. (a) 3.00 kV; (b) 750 V

5. (a) No. The electric field strength in between the plates is 2.5 × 106 V/m, which is reduced than the failure strength for air (3.0 × 106 V/m}); (b) 1.7 mm