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General Physics Lab (International Campus) Department of PHYSICS YONSEI University

Lab Manual

Capacitors and Capacitance Ver.20170921

Lab Office (Int’l Campus)

Room 301, Building 301 (Libertas Hall B), Yonsei University 85 Songdogwahak-ro, Yeonsu-gu, Incheon 21983, KOREA (☏ +82 32 749 3430) / 13

[International campus Lab]

Capacitors and Capacitance

Investigate the force between the charged plates of a parallel plate capacitor and determine the capacitance of the capacitor.

Determine the effect of a dielectric between the plates of a parallel plate capacitor.

1. Capacitors and Capacitance

A capacitor is a device that stores electric potential energy

and electric charge. Capacitors have a tremendous number

of practical applications in devices such as electronic flash

units for photography, microphones, and radio receivers.

Fig 1 Any two conductors 𝑎𝑎 and 𝑏𝑏 insulated from each

other form a capacitor.

Any two conductors separated by an insulator (or a vacuum)

form a capacitor (Fig.1). Each conductor initially has zero net

charge and electrons are transferred from one conductor to

the other; this is called charging the capacitor. Then the two

conductors have charges with equal magnitude and opposite

sign. When we say that a charge 𝑄𝑄 is stored on the capacitor,

we mean that the conductor at higher potential has charge

+𝑄𝑄 and the conductor at lower potential has charge −𝑄𝑄.

The electric field at any point in the region between the con-

ductors is proportional to the magnitude 𝑄𝑄 of charge on each

conductor. It follows that the potential difference 𝑉𝑉 between

the conductors is also proportional to 𝑄𝑄. If we double the

magnitude of charge on each conductor, the charge density

at each point doubles, the electric field at each point doubles,

and the potential difference between conductors doubles;

however, the ratio of charge to potential difference does not

change. This ratio is called the capacitance 𝐶𝐶 of the capaci-

tor:

𝐶𝐶 =𝑄𝑄𝑉𝑉 (1)

The greater the capacitance 𝐶𝐶 of a capacitor, the greater

the magnitude 𝑄𝑄 of charge on either conductor for a given

potential difference 𝑉𝑉 and hence the greater the amount of

Objective

Theory

----------------------------- Reference --------------------------

Young & Freedman, University Physics (14th ed.), Pearson, 2016

24.1 Capacitors and Capacitance (p.810~814)

24.4 Dielectrics (p.821~826)

24.2 Capacitors in Series and Parallel (p.814~817)

22.3 Gauss’s Law (p.753~757)

22.4 Applications of Gauss’s Law – Ex.22.8 (p.760~761)


Lab Manual




stored energy. Thus capacitance is a measure of the ability of

a capacitor to store energy. We will see that the value of the

capacitance depends only on the shapes and sizes of the

conductors and on the nature of the insulating material be-

tween them. The SI unit of capacitance is called farad (F).

1 F = 1 farad = 1 C/V = 1 coulomb/volt

The simplest form of capacitor consists of two parallel con-

ducting plates, each with area 𝐴𝐴, separated by a distance 𝑑𝑑

that is small in comparison with their dimension (Fig. 2). We

call this arrangement a parallel-plate capacitor. When the

plates are charged, because opposite charges attract, most

of the charge accumulates at the opposing faces of the plates

and the electric field is almost completely localized in the

region between the plates. A small amount of charge resides

on the outer surfaces of the plates, and there is some fringing

of the field at the edges (Fig. 3a). But if the plates are very

large in comparison to the distance between them, the

amount of charge on the outer surface is negligibly small, and

the fringing can be neglected except near the edges. In this

case we can assume that the field is uniform in the interior

region between the plates, as shown in Fig. 3b, and the

charges are distributed uniformly over the opposing surfaces.

Fig 2 A charged parallel-plate capacitor.

Fig 3 Electric field between oppositely charged parallel plates.

If the surface charge densities of each plate are +𝜎𝜎 and

−𝜎𝜎, we can find the electric field in the region between the

plates using following equation known as Gauss’s law.

Φ𝐸𝐸 = �𝐸𝐸⊥𝑑𝑑𝐴𝐴 = �𝑬𝑬��⃗ ⋅ 𝑑𝑑𝑨𝑨��⃗ =𝑄𝑄encl𝜖𝜖0

(2)

We consider a cylindrical Gaussian surface 𝑆𝑆1 with flat

ends of area 𝐴𝐴′ (Fig. 3b). The upper end of surface is within

the positive plate. Since the field is zero within the volume of

any solid conductor under electrostatic conditions, there is no

electric flux through this end. The electric field between the

plates is perpendicular to the lower end, so on that end, 𝐸𝐸⊥

is equal to 𝐸𝐸 and the flux is 𝐸𝐸𝐴𝐴′; this is positive, since 𝑬𝑬��⃗ is

directed out of the Gaussian surface. There is no flux through

the side walls of the cylinder, since these walls are parallel to

𝑬𝑬��⃗ . So the total flux integral in Gauss’s law is 𝐸𝐸𝐴𝐴′. The net

charge enclosed by the cylinder is 𝜎𝜎𝐴𝐴′, so equation (2) yields

𝐸𝐸𝐴𝐴′ = 𝜎𝜎𝐴𝐴′ 𝜖𝜖0⁄ ; we then have

𝐸𝐸 =𝜎𝜎𝜖𝜖0

(3)

The Gaussian surface 𝑆𝑆4 yields the same result. Surfaces

𝑆𝑆2 and 𝑆𝑆3 yield 𝐸𝐸 = 0 . The magnitude of surface charge

density 𝜎𝜎 is equal to the magnitude of the total charge 𝑄𝑄 on

each plate divided by the area 𝐴𝐴 of the plate, or 𝜎𝜎 = 𝑄𝑄/𝐴𝐴, so

equation (3) can be expressed as

𝐸𝐸 =𝜎𝜎𝜖𝜖0

=𝑄𝑄𝜖𝜖0𝐴𝐴

(4)

The field is uniform and the distance between the plates is

𝑑𝑑, so the potential difference between the two plates is

𝑉𝑉 = 𝐸𝐸𝑑𝑑 =1𝜖𝜖0𝑄𝑄𝑑𝑑𝐴𝐴 (5)

From equations (1) and (5), we see that the capacitance 𝐶𝐶

of a parallel-plate capacitor in vacuum is

𝐶𝐶 =𝑄𝑄𝑉𝑉 = 𝜖𝜖0

𝐴𝐴𝑑𝑑 (6)


Lab Manual




The capacitance depends only on the geometry of the ca-

pacitor; it is directly proportional to the area 𝐴𝐴 of each plate

and inversely proportional to their separation 𝑑𝑑. 𝜖𝜖0 is a uni-

versal constant.

𝜖𝜖0 = 8.85 × 10−12 C2/N ∙ m (𝑜𝑜𝑜𝑜 F/m)

If the fields due to the charges on each plate are 𝑬𝑬��⃗ 𝑎𝑎 and

𝑬𝑬��⃗ 𝑏𝑏 in Fig. 3b, the field between the plates is 𝑬𝑬��⃗ = 𝑬𝑬��⃗ 𝑎𝑎 + 𝑬𝑬��⃗ 𝑏𝑏

and 𝑬𝑬��⃗ 𝑎𝑎 = 𝑬𝑬��⃗ 𝑏𝑏 since the plates have the same magnitude but

opposite sign of total charge 𝑄𝑄. So the field with the charge

𝑄𝑄 on the upper plate is 𝐸𝐸𝑎𝑎 = (1 2⁄ )𝐸𝐸 = 𝜎𝜎 2𝜖𝜖0⁄ = 𝑄𝑄 2𝜖𝜖0𝐴𝐴⁄ and

the magnitude of the electrostatic force exerted on the charge

−𝑄𝑄 on the lower plate by the upper plate is given by

𝐹𝐹 = (−𝑄𝑄)𝐸𝐸𝑎𝑎 = −𝑄𝑄2

2𝜖𝜖0𝐴𝐴== −

𝜖𝜖0𝐴𝐴𝑉𝑉2

2𝑑𝑑2 (7)

The negative sign of this equation means that the attractive

force is exerted on the lower plate. From equations (5), (6)

and (7), we can express the capacitance 𝐶𝐶 of a capacitor as

a function of the force 𝐹𝐹 exerted on the lower plate.

𝐶𝐶 =2𝐹𝐹𝑑𝑑𝑉𝑉2 (8)

2. Dielectrics

Most capacitors have a dielectric between their conducting

plates. Placing a solid dielectric between the plates of a ca-

pacitor serves several functions, one of which is that the ca-

pacitance of a capacitor of given dimensions is greater when

there is a dielectric material between the plates than when

there is vacuum.

Consider a charged capacitor with magnitude of charge 𝑄𝑄

on each plate and potential difference 𝑉𝑉0. When we insert an

uncharged sheet of dielectric between the plates, the poten-

tial difference decreases to a smaller value 𝑉𝑉. When we re-

move the dielectric, the potential difference returns to its orig-

inal value 𝑉𝑉0.

The original capacitance 𝐶𝐶0 is given by 𝐶𝐶0 = 𝑄𝑄 𝑉𝑉0⁄ , and the

capacitance 𝐶𝐶 with the dielectric present is 𝐶𝐶 = 𝑄𝑄 𝑉𝑉⁄ . The

charge 𝑄𝑄 is same in both cases, and 𝑉𝑉 is less than 𝑉𝑉0, so

we conclude that the capacitance 𝐶𝐶 with the dielectric pre-

sent is greater than 𝐶𝐶0. When the space between plates is

completely filled by the electric, the ratio of 𝐶𝐶 to 𝐶𝐶0 is called

the dielectric constant 𝐾𝐾 of the material:

𝐾𝐾 =𝐶𝐶𝐶𝐶0

(9)

The dielectric constant 𝐾𝐾 is a pure number. Because 𝐶𝐶 is

always greater than 𝐶𝐶0, 𝐾𝐾 is always greater than unity. For

air at ordinary temperatures and pressures, 𝐾𝐾 = 1.00059.

Material 𝐾𝐾 Material 𝐾𝐾

Vacuum 1 Mica 3~6

Air (1 atm) 1.00059 Glass 5~10

Air (100 atm) 1.0548 Germanium 16

Teflon 2.1 Glycerin 42.5

Polyethylene 2.28 Water 80.4

When the charge is constant, 𝑄𝑄 = 𝐶𝐶0𝑉𝑉0 = 𝐶𝐶𝑉𝑉 and 𝐶𝐶 𝐶𝐶0⁄ =

𝑉𝑉0 𝑉𝑉⁄ . In this case, equation (9) can be rewritten as

𝑉𝑉 =𝑉𝑉0𝐾𝐾 (10)

When a dielectric material is inserted between the plates

while the charge is kept constant, the potential difference

between the plates decrease by a factor 𝐾𝐾. Therefore the

field between the plates must decrease by the same factor.

𝐸𝐸 =𝐸𝐸0𝐾𝐾 (11)

Fig 4 A common type of capacitor uses dielectric sheet to

separate the conductors.


Lab Manual




Since the electric-field magnitude is smaller when the dielec-

tric is present, the surface charge density must be smaller as

well. The surface charge on the conducting plates does not

change, but an induced charge of the opposite sign appears

on each surface of the dielectric (Fig. 5).

The dielectric was originally electrically neutral and is still

neutral; the induced surface charges arise as a result of re-

distribution of positive and negative charge within the dielec-

tric material, a phenomenon called polarization. The induced

surface charge is directly proportional to the electric-field

magnitude 𝐸𝐸 in the material for many common dielectrics

except special cases, so 𝐾𝐾 is a constant.

We can derive a relationship between this induced surface

charge and the charge on the plates. Let’s denote the magni-

tude of the charge per unit area induced on the surfaces of

the dielectric by 𝜎𝜎𝑖𝑖 . The magnitude of the surface charge

density on the capacitor plates is 𝜎𝜎. Then the net surface

charge on each side of the capacitor has magnitude (𝜎𝜎 − 𝜎𝜎𝑖𝑖).

Now we can use equation (4) to express the field without and

with dielectric respectively, then we have

𝐸𝐸0 =𝜎𝜎𝜖𝜖0

𝐸𝐸 =𝜎𝜎 − 𝜎𝜎𝑖𝑖𝜖𝜖0

(12)

Fig 5 (a) Electric field of magnitude 𝐸𝐸0 with vacuum between charged plates. (b) Introduction of a dielectric of dielectric constant 𝐾𝐾. (c) The induced surface charges and their field. (d) Resultant field of magnitude 𝐸𝐸0/𝐾𝐾. For a given charge density 𝜎𝜎, the induced charges on the dielectric’s surfaces reduce the electric field between the plates.

Using these expressions in equation (11) and rearranging

the result, we find

𝜎𝜎𝑖𝑖 = 𝜎𝜎 �1 −1𝐾𝐾�

(13)

This equation shows that when 𝐾𝐾 is very large, 𝜎𝜎𝑖𝑖 is nearly

as large as 𝜎𝜎, in this case, 𝜎𝜎𝑖𝑖 nearly cancel 𝜎𝜎, and the field

and potential difference are much smaller than their value in

vacuum, in other words, a capacitor can stores more electric

charges for a given 𝑉𝑉 when there is a dielectric material be-

tween the plates than when there is vacuum.

The product 𝐾𝐾𝜖𝜖0 is called the permittivity of the dielectric,

denoted by 𝜖𝜖:

𝜖𝜖 = 𝐾𝐾𝜖𝜖0 (14)

In terms of 𝜖𝜖 we can express the electric field within the die-

lectric as

𝐸𝐸 =𝜎𝜎𝜖𝜖 (15)

The capacitance when the dielectric is present is given by

𝐶𝐶 = 𝐾𝐾𝐶𝐶0 = 𝐾𝐾𝜖𝜖0𝐴𝐴𝑑𝑑 = 𝜖𝜖


In empty space, where 𝐾𝐾 = 1, 𝜖𝜖 = 𝜖𝜖0, equation (16) reduce

to equation (4) for a parallel-plate capacitor in vacuum. For

this reason, 𝜖𝜖0 is sometimes called the permittivity of vacu-

um. Equation (16) shows that extremely high capacitances 𝐶𝐶

can be obtained with plates that have a large surface area 𝐴𝐴

and are separated by a small distance 𝑑𝑑 by a dielectric with

a large value of 𝐾𝐾.


Lab Manual




1. List

Item(s) Qty. Description

Capacitor Apparatus

1 Constructs variable parallel plate capacitors.

High Voltage Power Supply (Power cord included)

1 Produces high voltage up to 25kV.

Patch Cords (High Voltage) (with safety shrouded banana plugs)

2 Carry high voltage power.

Capacitor Plates Set

1 set Constructs parallel plate capacitors. Plate 1: 𝑜𝑜 = 53 mm Plate 2: 𝑜𝑜 = 75 mm

Dielectrics

1 set Increase the capacitance a capacitor. Acryl: 𝑜𝑜 = 75 mm, 𝑑𝑑 = 3 mm, 𝐾𝐾 = 2.56 Glass: 𝑜𝑜 = 75 mm, 𝑑𝑑 = 3 mm, 𝐾𝐾 = 5.6

Electronic Balance (DC adaptor included)

1 Measures mass of an object with a precision to 0.01 g.

Discharger

1 Releases stored electric charge from capacitor plates.

Bubble Level

1 Checks the level of a surface.

Equipment


Lab Manual




2. Details

(1) Capacitor Apparatus

The Apparatus forms a parallel plate capacitor using various

sizes of plate electrodes. The micrometer mechanism allows

the spacing between the plates to be adjusted in small in-

crements, thus permitting the system to be tuned.

The spindle of the micrometer has one thread per millimeter,

and thus one completer revolution moves the spindle through

a distance 1 mm. The thimble has 100 graduations, each

being 0.01 mm. Thus, turning the thimble through one divi-

sion (1/100 turn) moves the spindle axially 0.01 mm.

(2) High voltage Power Supply

The high voltage power supply provides very high voltage

up to 25,000 V. The current available from the power supply

is too low to cause any permanent damage. However, the

voltage is high enough to cause a distinctly unpleasant sen-

sation. Do not touch any connectors or connected conduc-

tors while the power supply is turned on.

(3) Electronic Balance


Lab Manual




(1) Level the capacitor apparatus.

Using a bubble level as a reference, level the apparatus by

adjusting the leveling feet of the platform.

(2) Place the electronic balance on the platform.

(3) Mount the capacitor plates (𝑜𝑜 = 53mm) in the apparatus.

① Lift up the moving arm for easy installation of upper plate.

② Set the micrometer about 10 mm position.

③ Insert the upper plate into the holder of the moving arm.

④ Place the lower plate on the balance.

Lower Plate Upper Plate

Procedure

Caution

To prevent electric shock, keep these instructions in

mind. Although the current available from the power sup-

ply is too low to cause any permanent damage, the volt-

age on the capacitor plates is high enough to cause a

distinctly unpleasant sensation if you touch them.

1. Do not touch any plates or connectors while the pow-

er supply is turned on.

2. Prior to touching the plates or cables, you should be

sure to discharge the charges stored in the apparatus by

turning off the grounded power supply.

3. After turning off the power supply, you should always

double check if the charges remain on the plates. Use

the discharger to release any residual charge on the

plates by touching them at the same time.

4. If an arc occurs, as indicated by a sizzling noise, turn

off the power supply immediately and make sure the

plates are exactly parallel. It also could occur due to high

humidity. If it does, change your experimental conditions

i.e. increase the separation of the plates or decrease the

voltage.


Lab Manual




(4) Connect the power supply to the plates.

DO NOT turn on the power supply until you finish setup.

(5) Level the electrode plates.

① Lower the moving arm until the separation of the plates

becomes about 5 mm.

② Adjust the level of the lower plate using the level feet of

the balance. Make sure the plates are exactly parallel.

(6) Turn on the electronic balance.

Zero the balance by pressing [영점] or [용기] button. The

balance is sensitive to very small forces and vibrations. Avoid

touching the apparatus including connected patch cords while

making your measurements.

Caution

The electronic balance is a precision instrument. Sub-

jecting it to impact could cause it to fail. Treat it with care.

Note

For zeroing the balance, press [영점] and stand by until

the [ZERO] mark lights up on the display.

If the initial value is relatively high, [영점] button will not

work. In this case, you can zero it using [용기] button.

([TARE] mark will light up on the display.)

If you have any problem zeroing the balance, turn the

power off and then on again.

Note

If any of [CT], [%], [PCS], [CHK], or [ANI] symbols lights

up on the display, press [모드] repeatedly until all sym-

bols disappear.


Lab Manual




(7) Adjust the plate separation to 0 mm.

Using the micrometer, adjust the plate separation to 0 mm.

This is indicated by the mass reading increasing suddenly.

Record the micrometer reading as a reference point

𝑑𝑑 = 0 mm. (Make sure the plates are exactly parallel.)

(8) Adjust the plate separation to 6 mm.

Using the micrometer, slowly increase the plate separation

𝑑𝑑 to 6 mm.

(9) Set the potential difference 𝑉𝑉 across the parallel plates

capacitor to 4 kV.

With the voltage adjust knob set at zero, switch on the pow-

er supply and gradually increase the voltage until the voltme-

ter shows 4 kV.

(10) Record the mass 𝑚𝑚.

The balance will show negative values because an attractive

force is exerted on the lower plate.

(11) Calculate the force 𝐹𝐹 = 𝑚𝑚𝑚𝑚 exerted on the lower plates.

(12) Using equation (8), calculate the measured value of the

capacitance 𝐶𝐶 of your capacitor.

𝐶𝐶 =2𝐹𝐹𝑑𝑑𝑉𝑉2 (8)

(13) Using equation (6), calculate the theoretical value of the

capacitance 𝐶𝐶 of your capacitor.

𝐶𝐶 =𝑄𝑄𝑉𝑉 = 𝜖𝜖0


(14) Compare two values above by finding the percent error.

Caution

This micrometer features 0~25 mm measuring range.

Do not rotate the thimble more than the maximum scale

setting.

Caution

Do not touch any plates or connectors while the power

supply is turned on. Although the current from the power

supply is too low to cause any permanent damage, the

voltage on the capacitor plates is high enough to cause a

distinctly unpleasant sensation if you touch them.

Caution

If an arc occurs, turn off the power supply immediately.

1. Make sure the plates are exactly parallel.

2. If the balance malfunctions, turn the power off and

then on again.

An arc could occur due to high humidity. In this case,

change your experimental conditions, i.e. increase the

separation of the plates or decrease the voltages.


Lab Manual




(15) Repeat the steps (9)-(14) for each of following 𝑉𝑉.

Record the values obtained from three independent meas-

urements with each voltage and find the average values.

𝑑𝑑

(mm)

𝑉𝑉

(kV)

𝑚𝑚

(kg)

𝐹𝐹 = 𝑚𝑚𝑚𝑚

(N)

𝐶𝐶 = 2𝐹𝐹𝑑𝑑 𝑉𝑉2⁄

(pF = 10−12 F)

6

4

6

8

10

𝑑𝑑

(mm)

𝑜𝑜

(m)

𝐴𝐴 = 𝜋𝜋𝑜𝑜2

(m2)

𝐶𝐶theory = 𝜖𝜖0𝐴𝐴 𝑑𝑑⁄

(pF = 10−12 F)

6 0.053

Verify the capacitance is always constant if the shape and

size of the capacitor does not change.

(16) Repeat your experiments for each of following 𝑑𝑑.

Set 𝑉𝑉 = 8 kV and record the values obtained from three in-

dependent measurements with each separation and find the

average values.

𝑑𝑑

(mm)

𝑉𝑉

(kV)

𝑚𝑚

(kg)


(N)


(pF = 10−12 F)

6

8

8

10

12

𝑑𝑑

(mm)

𝑜𝑜

(m)


(m2)


(pF = 10−12 F)

6

0.053

8

10

12

Plot a graph of 𝐶𝐶 as ordinates against the matching values

of 1/𝑑𝑑 as abscissa and verify 𝐶𝐶 is proportional to 1/𝑑𝑑.

(17) Repeat step (16) for the different area 𝐴𝐴 of the plates.

Mount the capacitor plates of radius 𝑜𝑜 = 75 mm. The area

𝐴𝐴 of the 𝑜𝑜 = 75 mm plate is twice as large as that of the

𝑜𝑜 = 53 mm plate.

𝑑𝑑

(mm)

𝑉𝑉

(kV)

𝑚𝑚

(kg)


(N)


(pF = 10−12 F)

6

8

8

10

12

𝑑𝑑

(mm)

𝑜𝑜

(m)


(m2)


(pF = 10−12 F)

6

0.075

8

10

12

Caution

Before touching the plates or cables, you should be sure

to discharge the charges stored in the apparatus by turn-

ing off the power supply and touching the discharger.


Lab Manual




(18) Compare the results of step (16) and (17).

Find the relationship between 𝐶𝐶 and 𝐴𝐴.

(19) Repeat your experiments using dielectrics.

Mount 𝑜𝑜 = 75mm capacitor plates and adjust the plate sep-

aration 𝑑𝑑 to 6mm . Insert acryl or glass dielectric having

thickness 3mm between the plates. Set 𝑉𝑉 = 6 kV.

Record the values obtained from three independent meas-

urements with each dielectric and find the average values.

Dielectric

substance

𝑑𝑑

(mm)

𝑉𝑉

(kV)

𝑚𝑚

(kg)


(N)

𝐶𝐶

(pF = 10−12 F)

only air 6

6

acryl + air 3 + 3

glass + air 3 + 3

Dielectric

substance

𝑑𝑑

(mm)

𝑜𝑜

(m)


(m2)

𝐶𝐶theory

(pF = 10−12 F)

only air 6

0.075

acryl + air 3 + 3

glass + air 3 + 3

* Reference:

- Acryl: 𝐾𝐾 = 2.56

- Glass: 𝐾𝐾 = 5.6

Caution

Before touching the plates or cables, you should be sure

to discharge the charges stored in the apparatus by turn-

ing off the power supply and touching the discharger.

Caution

Handle the glass dielectric with care. It is very fragile.

Note

If a dielectric with thickness 𝑑𝑑′ = 3mm is inserted be-

tween the plates separated by a distance 𝑑𝑑 = 6mm, then

this capacitor is equivalent to two capacitors in series as

shown below.

Suppose the capacitances of each capacitor are 𝐶𝐶1

and 𝐶𝐶2 , the dielectric constants are 𝐾𝐾1 and 𝐾𝐾2 , and

each is separated by a distance 𝑑𝑑′ = 𝑑𝑑1 = 𝑑𝑑2 = (1 2⁄ )𝑑𝑑,

then

1𝐶𝐶 =

1𝐶𝐶1

+1𝐶𝐶2

or 𝐶𝐶 =𝐶𝐶1𝐶𝐶2𝐶𝐶1 + 𝐶𝐶2

𝐶𝐶1 = 𝐾𝐾1𝜖𝜖0𝐴𝐴𝑑𝑑′ and 𝐶𝐶2 = 𝐾𝐾2𝜖𝜖0

𝐴𝐴𝑑𝑑′

𝐶𝐶 = �𝐾𝐾1𝐾𝐾2𝐾𝐾1 + 𝐾𝐾2

� 𝜖𝜖0𝐴𝐴𝑑𝑑′


Lab Manual




Your TA will inform you of the guidelines for writing the laboratory report during the lecture.

Please put your equipment in order as shown below.

□ Delete your data files and empty the trash can from the lab computer.

□ Turn off the Computer.

□ With the voltage control knob set at zero, turn off the Power Supply and unplug the power cable.

□ Turn off the Electronic Balance and unplug the dc adaptor.

□ Place the Electronic Balance on the platform of the capacitor apparatus.

□ Put the Capacitor Plates, Dielectric Plates, Patch Cords, Bubble Level, and Discharger in the container.

□ Handle the glass Dielectric Plate with care. It is very fragile.

Result & Discussion

End of Lab Checklist

objective theory - yonsei universityphylab.yonsei.ac.kr/exp_ref/202_capacitor_eng.pdf ·...

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