HIGH VOLTAGE TECHNOLOGY MODULE (QUESTION AND ANSWER).docx

HIGH VOLTAGE TECHNOLOGY QUESTION AND ANSWER

PREPARED BY

ASSOC. PROF. DR. ZOLKAFLE BIN BUNTAT

MAY 2013

CHAPTER 1 ELECTRIC FIELDS (QUESTION & ANSWER)

Question 1 a. There are two types of field distribution, known as homogeneous and non-homogeneous field. What are the differences of both field distributions? State the electrodes-gap configuration to simulate the homogeneous and non-homogeneous field. Answer: i.

Homogeneous field 

E is the same throughout the field region.



Uniform or approximate uniform field distributions exists between two infinite parallel plates, or 2 spheres of equal diameters with gap spacing < sphere radius.



“Profiled” parallel plates of finite sizes are also used to simulate homogeneous field.

ii.

Nonhomogeneous field 

E is different at different points in the field region.



In the absence of space charges, E usually obtains the maximum value at the surface of the conductor which has the smallest radius of curvature – nonhomogeneous and asymmetrical.



Most of the practical HV components have nonhomogeneous and asymmetrical field distribution..



In some gaps – will produce nonhomogeneous fields and symmetrical, e.g. rod-rod or sphere-sphere (large distance between spheres) gaps.



HV electrode has higher E than the grounded electrode.

b. Experimental analogue is one of the methods for determining the potential distribution. Briefly describe any two of the experimental analogs used for space-charge-free fields. Answer: i.

Electrolytic Tank 

Widely used for decades.



Equipotential boundaries are represented in the tank by specially form sheets of metal.



Example, a single dielectric problem such as a three-core cable may be represented by electrolytes of different conductivities separated by special partitions.

ii.

Semiconducting Paper Analog 

Less accurate but attractively simple alternative to the electrolyte.



Errors from this method result from the non-homogeneity of the paper resistivity.



Errors also dependence on the ambient humidity and the contact resistance to the electrodes.

iii.

Resistive Mesh Analog 

The continuous field is replacing by a discrete set of points as depicted by a mesh of resistors.



The used of discrete resistors introduce an error arising from the finite mesh analysis.



This error may be reduced by reducing the mesh size.

c. A high voltage DC transmission line rated at 132 kV peak traverses a location where a road shall be constructed below it as shown in Figure Q1. The metallic walls L1 and L2 are energized at 20 kV peak and 80Kv respectively, and each standing on the insulator made of polycarbonate. Use the Finite Difference Method to determine the potential at point 2, 4 and 6. Limit the iteration process to two only.

Figure Q1

Answer:

1st iteration

2nd iteration

The potential at point 2, 4 and 6 after 2 iterations are; V2 = 65.53 kV,

V4 = 21.63 kV,

V6 = 47.73 kV

Question 2 a. Briefly describe any two (2) of the followings:

i.

Field enhancement factor

ii.

„Medium High Voltage‟ (MHV), „High voltage‟ (HV), „Extra high Voltage‟ (EHV) and „Ultra high voltage‟ (UHV)

iii.

Three applications of high voltages excluding those in the generation, transmission and distribution of electrical energy

Answer:

i.

Whereas any designer of the high voltage apparatus must have a complete knowledge of the electric field distribution, for a user of the system the knowledge of

the

maximum

value

of

the

electric

field

to which the insulation is likely to be subjected and the location of such maximum gradient point is generally sufficient. Consequently, the concept of field enhancement factor or simply field factor f is of considerable use. This factor is defined as

Where

the average is field in the gap and is equal to the applied potential/gap

separation between the electrodes

Field utilization factor

(larger

represents a more compact equipment)

ii.

iii.

The following voltage classification describes the meanings: Voltage class

Voltage range

Medium high voltage (MHV)

1kV= 300 kVrms. The level of voltage supply from customer side to the generation side can be categorized as i) Low Voltage (LV) ii) Medium Voltage (MV) iii) High Voltage (HV) iv) Extra High Voltage (EHV) v) Ultra High Voltage (UHV) These systems are affected by transient overvoltage generally due to external source such as lightning phenomenon and internally generated is the switching activities. In order system components can function as they should, they have to be design to withstand against lightning related surges as well as switching activities. Related to withstand on lightning surge BIL is to be referred and BSL for switching. System working more than 300kV BSL is more significant than BIL. (c) i) 220kV ii) P = 16%, V = 211.2kV P = 84%, V = 228.1kV Standard Deviation = V50 – V16 = 220 – 211.8 = 8.2kV Standard Deviation = V84 – V50 = 228.1 – 220 = 8.1kV Coefficient of variation = 8.1/220 = 0.037 pu iii) Statistical withstand voltage = (1 – 3 Coeff)V50 = (1 – 3*0.037)*220

= 195.6kV iv) Statistical flashover voltage = (1 + 3 Coeff)V50 = (1 + 3*0.037)*220 = 244.4Kv

100 90 84%

80

Percentage (%)

70 60 50 40 30 20 16%

10 0 200

210

220

230

240

250

Breakdown Voltage (kV)

Question 7: a) Discuss three (3) mechanisms of solid insulation breakdown.

260

Answer Q7:

1.

Intrinsic Breakdown 

When the high voltage applied for a long time.



Electrical power is determined by the intrinsic strength.



Depend on the presence of free electrons that move through the lattice of solid.



2.

These electrons produce a current flow.

Electromechanical Breakdown 

Less rigid insulation material (rubber, PVC).



Electrostatic forces that act exceed the strength of solid mechanics.



Mechanical damage will occur.

3.

Thermal Breakdown 

Leakage current flow when the electric stress applied.



Solid temperature increased by the heating process.



Heat transferred to the insulation surroundings by conduction and radiation process.



4.

Limiting the maximum thickness of a solid.

Chemical And Electrochemical Breakdown 

Chemical changes if the electrical stress is continuously react with air and gas.



Reactions that occur - oxidation, hydrolysis, chemical reaction.



Can be reduced by checking the material carefully.

5.

Treeing And Tracking Breakdown 

Two effects when electric stress is long: i.

the presence of conductive path across the surface.

ii.

effects of leakage current due to sparks through the path that produces sparks.



During

the

process

oftracking,the

spread

ofsparksin

the

form

ofbranchescalledtreeing.

6.

Internal Discharge Breakdown 

Cavitycontains airinthe solidinsulation.



Fieldin the cavity larger than the field in theinsulation.



Breakdown exists in the cavity.

Question 8 :

a) Explain what is meant by the B.I.L of a high voltage equipment. Answer Q8: a) For equipment rated at less than 300 kV, it is a statement of the Basic Lightning Impulse Insulation Level (BIL) and the short duration power frequency withstand voltage.

Question 9 :

a) Describe in details with the appropriate diagrams, the methods of insulation coordination for over voltages.

b) In transmission system, components whose failure would have very severe consequences are often protected by protective devices. Describe the coordination of protected insulation with regards to a high voltage transformer.

Answer Q9: a) There are two methods of insulation coordination for overvoltages:

i) Conventional Method. This is the method used when the probability of failure for a given overvoltage is unknown, ie. With non-self restoring insulation. The known probability distribution of overvoltage amplitudes is used only to determine the maximum surge which is liable to occur. To this is added a safety margin to allow for the unknown flashover probability distribution. The resultant voltage level is specified as a „withstand‟ voltage,

which the insulation must be able to hold off.

ii) Statistical Method This is based on knowledge of the statistics of flashover probability. The design is then based on an acceptable risk of flashover. If at a voltage level V, the probability of failure is P(V) and the frequencies of occurrence or surges of that level is f(V) then the risk function is defined as, r(V) = P(V).f(V). The risk function is shown below,

For a given insulation design the total risk of failure will be:

∫

The risk is therefore determined by the area under the r(V) curve. As the insulation is strengthened by using larger gap or longer insulation strength, the risk of failure diminishes as in the diagram below.

For a chosen risk of failure it is therefore possible to position the curve. We can then choose a reference probability and quote the corresponding voltage as a statistical

withstands voltage 0.1, ie.

. In practice the reference probability is normally taken as P(V) =

is the voltage at which the insulation has a 90% probability of withstand.

The system design engineer will therefore specify that the insulation must have a withstand of

based on the acceptable risk criterion, and the insulation design

engineer will develop insulation suitable for that withstand level.

b) In transmission systems, component whose failure would have very severe consequences are often protected by devices such as surge diverters or spark gaps. These devices are designed to break down in preference to the insulation of the equipment to be protected such as high-voltage transformer. The coordination of the protective devices insulation with regards to the protected equipment can be made by using the conventional and statistical methods. In applying the conventional method, the maximum allowable overvoltage is coordinated with the protective level of the device, ie. The protective device is set to flashover if

for the more expensive equipments is

exceeded.

In applying the statistical method the flashover probabilities

(V) and P(V) of the

protective device and the equipment which it protects are used to calculate a probability P*(V) that the main insulation will flash over in spite of the protective device. The risk is then found form R*=∫

If the protective device is a spark gap, it itself will be dimensioned by normal methods to give a risk of flashover which although greater than other devices, is still limited. If surge diverters are used there is no reason to limit the risk of flashover as surge diverters automatically restore the system voltage due to their non-liner resistance characteristics without damage. However if a surge diverter is triggered by a high-voltage switching surge or lightning transient.

Question 10:

(a) Describe the secondary processes which can lead to an electron avalanche and how these processes may be identified. Show that the discharge current in a multi avalanche Townsend process in a non-attaching gas is given by

] Where

– initial voltage

Α – first Townsend ionization coefficient - Second Townsend ionization coefficient d – Gap distance in cm

(b) Measurement of breakdown voltages in a uniform field spark gap in air gave the result as shown in the Table Q10 Table Q10 Breakdown Gap spacing (mm)

Pressure (Bar)

Temperature ( )

2.5

1.03

30

0.91

27

1.18

15

88.38

Use the expression derived from Paschen‟s law i)

voltage,

(√

) determine

The relative air density , referred to standard atmospheric conditions of 1013.25 mbar and 20˚C.

ii) The value of constant A and B. iii) The breakdown voltage of a 3cm gap spacing at pressure of 3000 mbar and a temperature of 20˚C.

Answer Q10:

a) The electrical breakdown of a gas is brought about by various processes of ionization. These gas processes involving the oscillation of electrons, ions and photons with gas molecules and electrode processes which take place at or near the electrode surface. When a pair of electrodes is immersed in a gas and a voltage applied across them the current voltage characteristics of figure below is observed Io

Self-sustain I

I Non-Self-

Io

Ionization V

Charge collectio

At low voltage the observe current is due to collection of free carriers in the gap and as the voltage is increased a level is reached at which the free electrons gain enough energy to ionize. Electrons produced may cause further ionization so that an electron avalanche is generated. Ionization is the process by which an electron is removed from an atom, leaving the atom with a net positive charge. The probability of ionization due to the electron will depend on the number of collision made per unit distance with coefficient referred as the primary ionization coefficient which is the number of ionization collisions per unit electron per cm travel. With the primary ionization alone the discharged is not self sustaining. If the source of initial electrons is removed the current, the current I falls to zero. This suggests that processes other than simple - process are occurring. This additional current is produced by secondary emission processes. A secondary ionization coefficient

is defined as the number of secondary electrons

produced at the cathode produced in the gap These processes for secondary electrons liberation can be identified by: i) Positive ion

- ions do not have enough energy to ionize gas molecules directly but

may release electrons on colliding with the cathode surface

ii) Photons

- a proportion of the collision in the gap cause excitation of neutral gas

molecules which in return to the ground state may be emit photons which release electrons by photoemission. iii) Metastables

- metastables molecules may diffuse to the cathode and release

electrons One or more secondary mechanism may exist giving a total secondary effect described

Let

= number of initials electrons at cathode = The number of secondaries = the total emission including secondaries i.e

Self-sustain Vs s

V Charge harge arge rge ge e collection

Ionization Breakdown

At The total number of new electros produced,

ollectionI Non-Self-sustain

If

electrons are produced at the cathode per ionizing collision in the gap, then

Thus,

Under steady state conditions, I=

Breakdown

Gap spacing (mm)

Pressure (Bar)

Temperature ( )

2.5

1.034

30

0.91

27

1.180

15

88.38

voltage,

Question 11:

a)

What is the basic different between self-restoring and non-self-restoring insulation?

b)

Describe the differences between Basic Lightning Impulse Insulation Level (BIL) and Basic Switching Impulse Insulation Level (BSL)

c)

Name the types of BIL that are mentioned in the standards? (How many type of Basic Lightning Impulse Insulation Level and name them?)

d)

Name the types of BSL that are mentioned in the standards? (How many type of Basic Switching Impulse Insulation Level and name them?)

e)

In relation to BIL and BSL, standards provide two types of tests to determine the best method to obtain BIL and BSL. (i) Explain those methods as provided by the standards document. (ii) Discuss which method is the best by not neglecting to discuss points for instance probability of passing test, probability of flashover per-impulse, manufacturer‟s risk and ideal test and other relevant factors.

Answer Q11:

a) Self-Restoring Insulation : Insulation that completely recovers insulating properties after a disruptive discharge (flashover) caused by the application of a voltage. Non Self -Restoring Insulation : Insulation that losses insulating properties after a disruptive discharge (flashover) caused by the application of a voltage.

b) For equipment rated at less than 300 kV, it is a statement of the Basic Lightning Impulse Insulation Level (BIL) and the short duration power frequency withstand voltage. For equipment rated at greater than 300 kV, it is a statement of the Basic Switching impulse Insulation Level (BSL) and the power frequency withstand voltage.

c) 2 type, statistical and conventional

d) 2 type, statistical and conventional

e) i) 1) The n/m test : m impulses are applied. The test is passed if no more than n result in flashover. 2) The n + m test : n impulses are applied. If none result in flashover, the test is passed. If there are two or more flashover, the test is failed. If only one flashover occurs, m additional impulses are applied and the test is passed if none of these result in a flashover.

ii) These alternate tests can be analysed statistically to determine their characteristic. That is a plot is constructed of the probability of passing the test as a function of the actual but unknown probability of flashover per application of a single impulse. The characteristics for the above three tests are shown in Fig 1.1

Question 12: a) Discuss three (3) mechanisms of solid insulation breakdown b) Show that the breakdown criterion in gas according to Paschen‟s Law is given by :

g

{

*

(

)

-

where,

- gap distance at sparkover voltage p-pressure - sparkover voltage f&g-different functions

c)

In nitrogen gas, the static breakdown voltage

of a uniform field gap may be

expressed as,

Where A and B are constants, p is the gas pressure in torr referred to a temperature of and d is the gap legth in cm. A 1 cm uniform field gap in nitrogen at 760 torr and

is found to breakdown of 33.3

kV. The pressure is then reduced and after a period of stabilization, the temperature and pressure are measured as

and 500 torr respectively. The breakdown voltage is

found to be reduce to 21.9 kV. If the pressure is further reduced to 350 torr while the temperature of the closed vessel is raised to

and the gap distance is increased to 2

cm, determine the breakdown voltage.

Answer Q12: a) Three (3) mechanisms of solid insulation breakdown :i) Intrinsic/Ionic breakdown ii) Electromechanical breakdown iii) Thermal breakdown

Intrinsic/Ionic breakdown.  Occurs at a very short duration of HV applied (10-8 s).

 Depends on the presence of free electron, which capable of migration thru the lattice of the dielectric. G CREATIVE AND INNOVATIVE MIND Electromechanical breakdown.

 Due to electrostatic compressive forces that exceed mechanical compressive strength.  The highest apparent electric stress before breakdown, if the thickness of specimen do  is compressed to d under applied HV;

 Mechanical instability occurs when d/do = 0.6, and Y : Young‟s modulus and  depends on mechanical stress.

Thermal breakdown  Conduction current flows – heats up the specimen and the temperature rises.  Heat generated transfers to the surrounding medium by conduction and radiation.  Breakdown occurs when heat generated > heat dissipated.  Heat generated is proportional to the frequency – thermal breakdown is more serious at high frequency.  Thermal breakdown stresses are lower under a.c. condition then d.c.

b) By neglectingattachment, breakdown criterion;

…………… (1) Since (Paschen‟s Law) ,

and

where f &g signify different function.

………………. (2)

At breakdown,

…….……….... (3)

=

Substitute (2) &(3) into (1) gives

(

),

*

(

)+

-

proved

c) Corrected pressure to standard temperature of

,

√

…….(1) , Corrected pressure to standard temperature of 20˚C

√

……………..(2) ………(3) …...….(4) (3)-(4):

Corrected pressure to standard temperature of

√

=27.66kV

,

CHAPTER 5 INSULATION DIAGNOSTIC & PARTIAL DISCHARGES (QUESTION AND ANSWER)

Question 1 a)

The dissipation factor or loss tangent is an indication to determine the performance of insulating property of insulators. The most common method used to determine the lost tangent is by using a.c bridges. i)

Sketch the circuit diagram of a high voltage Schering bridge for the measurement of loss tangent (tan ).

ii)

Derive the expression for tan

of the unknown series model of tge tested

sample when the high voltage standard capacitor used in the Schering Bridge is a loss-free type.

Answer i)

C : Capacitance of the sample r : Resistance of the sample : Standard capacitor ,

ii)

and

: Variable components of Schering bridge

At balance condition ;

; [

][

]

By equating the real and imaginary parts of both side; Therefore;

For series model, tan δ = 𝝎Cr, (

b)

)(

)

Breakdown in solid or liquid dielectrics arise from the action of the electrical discharges in internal gaseous cavitties. Draw an equivalent circuit for such an arrangement and derive the expression for energy dissipated in the cavity in one discharge.

Answer

d

-

Distan ce

o

Volts

d n0

Void in solid dielectric material

Equivalent circuit of void in the structure

Energy dissipated in the cavity in one discharge;

(

)

Where

Therefore

(

)(

(

c)

)

)

A solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric constant of 3. It has an embedded air filled void of 2.5mm diameter and 1mm thickness. It is subjected to a voltage of 100k Vrms. Find the voltage at which an internal discharge in the void can occur. The breakdown strength of air is 3kV/mm. Also determine the charge value each time there is discharge inside the void and what will be the value of charge as measured on the detector?

Answer The inception voltage of the stressed void Vi is given by: ( Where

⁄

⁄

)

- breakdown stress of the void = 3 kV/mm - thickness of the specimen = 10mm - thickness of the void = 1mm - dielectric constant of the specimen = 3

when = 1mm (

⁄

⁄

)=

Capacitance of void

When t = 1mm, area of void = π (2.5 x 10ˉ³)² / 4 = 491 x 10ˉ⁶ m² ˉ²

ˉ⁶

ˉ

ˉ³

Capacitance of slab

ˉ²

²

ˉ²

5.79 x 10ˉ ² F

The measured change (

ˉ )

Therefore, ˉ²

ˉ ˉ²

Question 2 a)

Using the circuit for series and parallel models of an insulating sample, derive the loss tangent (tan ) equation in terms of the capacitance (C) and resistance (R) of the sample.

Answer R4

r

b1

C4

C2 c

R3

b2

C a Solid Insulator Model Series Model Rs = Cs = Tan δ =

=

= ωCsRs

Va

I Vc

Vr

Cs

Vc

Ca

Rs t void

Solid Insulator Model Parallel Model

b)

What are partial discharges? Describe some of the typical partial discharges.

Answer Some of the typical partial discharges are: i)

Corona or gas discharge. These occur due to non-uniform field in sharp edges of the conductor subjected to high voltage especially when the insulation provided is air or gas or liquid. [Fig. (a)]

ii)

Surface discharges and discharges in laminated materials on the interfaces of different dielectric material such as gas/solid interface. [Fig. (a) and (b)]

iii)

Cavity discharges: When cavities are formed in solid or liquid insulating materials the gas in the cavity is over stressed and discharges are formed. [Fig. (d)]

iv)

Treeing Channels: High intensity fields are produced in an insulating material at its sharp edges and this deteriorates the insulating material. The continuous partial discharges so produced are known as treeing channels. [Fig. (e)]

Cb

(a )

(b )

(c )

(d )

c)

(e )

A sample of impregnated paper (Ԑᵣ=3.5) of thickness 2mm is placed between large parallel-plate electrodes. A cubical air-filled cavity with 2x2 mm length of its edges and 0.05mm deep, with its axis at right angles to the electrodes is located in the centre of the dielectric. A sinusoidal alternating voltage is applied between the electrodes. The breakdown voltages of air at different pd level is shown in figure Q5 (i)

Determine the discharge inception stress in kV/mm in the dielectric for a pressure of 550mm Hg.

(ii)

Calculate the apparent discharge magnitude.

(iii)

If the pressure in the void is doubled, determine the new inception voltage.

Answer i)

Discharge inception voltage, Vᵢ,

U

(

) U

where U : breakdown voltage of the void

*

(

)+ U

[

(

)] U

(

[

)]

U

U is taken from the graph At pd = (500)(0.05) = 25; therefore U≈0.46kV

Vi = 12.14 (0.46) = 5.58kV Inception stress in kV/mm Ei = Vi/d = 5.58/2 = 2.79kV/mm

ii)

Apparent discharge qₐ; ]

qₐ = δVₐ[

since Cb t; (

)

)

proven

In term of electric field strength and

Hence

(

)

And (

)

and

Electric stress in the void is greater than the dielectric stress across the sample. Partial discharge occurs due to the very small gap of the void even at the normal service voltage.

Question 4 a)

The dissipation factor or loss tangent is an indication to determine the performance of insulating property of insulators. The most common method used to determine the loss tangent is by using a.c. bridges. i)

Sketch the circuit diagram of a high voltage Schering bridge for the measurement of loss tangent (tan )

ii)

Derive the expression for tan

of the unknown series model of the tested sample

when the high voltage standard capacitor used in the Schering bridge is a lossfree type.

Answer

C4 a

Sample Standard with losses r

R

r2

R

b2

AC Supply

C: capacitance of the sample r: resistance of the sample

C2: standard capacitor R3, R4 and C4: variable components of Schering Bridge

ii) Given;

At balance condition - equation (i)

(

)(

)

(

)

By equating the real and imaginary parts of both side; Therefore:

; and

For series model,

, (

b)

)(

)

A solid insulating material of relative permittivity,

comprises a cylindrical air-filled

cavity of depth t. The material has thickness T, where its value is much greater compared to the depth of the cavity. Show that the voltage across the sample (material),

is given by the expression.

Where

is the voltage across the cavity.

From the above equation, explain why partial discharge can occur in the cavity even though only normal service voltage is applied across the insulating material.

Answer

T

Cb

t

bc1 void

Cc

Ca

Va V Va

Void in solid dielectric material

Equivalent circuit of void in the structure

Void capacitance,

(i)

Capacitance series with void,

(ii)

Remaining capacitance,

(iii)

Voltage across cavity,

(iv)

Therefore (

) (

)

Since T >> t; (

) (

(

)

)

Yields

proven

In term of electric field strength

and

Hence

(

)

And ( Since

)

, the electrical stress in the void is greater than the electrical stress across

the sample. Thus, partial discharge occurs due to the very small gap of the void even at the normal service voltage.

Question 5

a)

Partial discharge detection and measurements were limited to the laboratory due to high levels of electrical noise at the switchyards. With the aid of proper diagrams describe the main sources of interferences or noise which hampered the partial discharge detection process and the techniques to suppress the interferences.

Answer Typical noise source : i)

Power supply

ii)

Voltage regulator

iii)

High voltage source

iv)

Filtering of the HV source

v)

Feeder line and electrodes

vi)

Coupling capacitor

vii)

Loose conductive objects in the vicinity of the last location

viii)

Pulse shaped interferences

ix)

Electromagnetic waves by radio transmitter (harmonic interferences)

x)

Interference currents in ground system of PD measuring instrument

The interferences can be reduced by;

b)

i)

Filter grounding

ii)

Shielded room

iii)

Separate source

iv)

Filter AC source (Harmonics)

A solid dielectric has a small cavity. Draw an equivalent circuit for such an arrangement and define all symbols. Derive the expression for electric field strength across the cavity.

Answer

R4

AC Sup ply

C

r

Sa mpl e

C2 C4

R3

c

b2 b a

Void in solid dielectric material

Equivalent circuit of void in the structure

In the absence of the void, the electric field within the insulation would everywhere be Eₐ = Vₐ/d. Neglecting conductance, the insulation, which its void, can be represented by the three capacitances a, b, and c. The void capacitance is represented by c. If the void has length d1 and the cross sectional area A1 (perpendicular to d1) then;

C=

₁ ₁

The void is connected to the conducting plates through two capacitance b1 and b2 . Their series combination is represented by the single capacitance b. Clearly,

b=

₁ ₁

Where Ԑᵣ is the relative permittivity of the insulating material. The remaining capacitor a has the value,

Where A is the cross sectional area of the insulation, minus the (usually small) cross sectional area of the void.

The voltage across the void is;

And substituting for the capacitances gives, ₁ ₁

₁

In the term of electric field strength we have, Vₐ =Eₐ.T

and

Vc = EC.t

Where Ec is the dielectric field strength in the void. Substituting in equation gives, ₁ ₁

₁

Usually the void will be small, so that d₁