× π
π − −
−6 10 12 1.52 10 4
36 4 1000 10
= 10 –6/(0.25 × 10–13)
= 40 × 106 V/m = 40 MV/m = 400 kV/cm ψ = E.r = 600 kV [using equations 4.2 and 4.4]
C = 4πe0err=1.111µµF with r = 0.01.
Example 4.2. The field strength on the surface of a sphere of 1 cm radius is equal to the corona-inception gradient in air of 30 kV/cm. Find the charge on the sphere.
Solution. 30 kV/cm = 3000 kV/m = 3 × 106 V/m.
3 × 106 =
× ×
× π
π −12 −4
0 10 10
36 4 1000 Q
= 9 × 1013Q0
giving Q0 = 3.33 × 10–8 coulomb = 0.033 µC.
The potential of the sphere is
V = 3 × 106 × 10–2 = 30 kV.
4.2 FIELD OF SPHERE GAP
A sphere-sphere gap is used in h.v. laboratories for measurement of extra high voltages and for calibrating other measuring apparatus. If the gap spacing is less than the sphere radius, the field is quite well determined and the sphere-gap breaks down consistently at the same voltage with a dispersion not exceeding ±3%. This is the accuracy of such a measuring gap, if other precautions are taken suitably such as no collection of dust or proximity of other grounded objects close by. The sphere-gap problem also illustrates the method of successive images used in electrostatics.
Figure 4.2 shows two spheres of radii R separated by a centre-centre distance of S, with one sphere at zero potential (usually grounded) and the other held at a potential V. Since both spheres are metallic, their surfaces are equipotentials. In order to achieve this, it requires a set of infinite number of charges, positive inside the left sphere at potential V and negative inside the right which is held at zero potential. The magnitude and position of these charges will be determined from which the voltage gradient resulting on the surfaces of the sphere on a line joining the centres can be determined. If this exceeds the critical disruptive voltage, a spark break-down will occur. The voltage required is the breakdown voltage.
Fig. 4.2 The sphere gap.
Consider two point charges Q1 and Q2 located with a separation D, Figure 4.3. At a point P(x, y) with coordinates measured from Q1, the potentials are as follows:
Potential at P due to Q1 = 1 2 2
Fig. 4.3 Point charge Q1 and sphere of radius R.
The total potential at P is VP = ( / / ) This clearly shows that Q1 and Q2 must be of opposite polarity.
From (4.7), r22/r12 = ( ) / 2, This is an equation to a circle in the two-dimensional plane and is a sphere in three-dimensional space.
R = D(Q2/Q1)/{1−(Q2/Q1)2} ...(4.9) This requires Q2 to be less than Q1 if the denominator is to be positive. The centre of the zero-potential surface is located at (S1, 0), where
S1 = R
This makes S2 = R Q Q Q Q
Q Q D D
S
1 2 1 2
2 1 2 1 2
) / ( 1
) /
( =
= −
− ...(4.11)
Therefore, the magnitude of Q2 in relation to Q1 is Q2 =
1 2 1
1 S
Q R R
Q S = ...(4.12)
Also, S1S2 = R2 ...(4.13)
These relations give the following important property:
'Given a positive charge Q1 and a sphere of radius R, with Q1 located external to the sphere, whose centre is at a distance S1 from Q1, the sphere can be made to have a zero potential on its surface if a charge of opposite polarity and magnitude Q2 = (Q1R/S1) is placed at a distance S2 = R2/S1 from the centre of the given sphere towards Q1.'
Example 4.3. A charge of 10 µC is placed at a distance of 2 metres from the centre of a sphere of radius 0.5 metre (1-metre diameter sphere). Calculate the magnitude, polarity, and location of a point charge Q2 which will make the sphere at zero potential.
Solution. R = 0.5, S1 = 2 ∴ S2 = R2/S1 = 0.125 m Q2 = Q1R/S1 =10×0.5/2=2.5µC
The charge Q2 is of opposite sign to Q1. Figure 4.4 shows the sphere, Q1 and Q2.
Example 4.4. An isolated sphere in air has a potential V and radius R. Calculate the charge to be placed at its centre to make the surface of the sphere an equipotential.
Solution. From equation (4.4), Q = 4πe0VR. S2 = 0.125 m = R2/S
Q2 = – 2.5 µc = Q1R/S
Fig. 4.4 Location of image charge Q2 inside sphere to make sphere potential zero.
We now are in a position to analyze the system of charges required to make one sphere at potential V and a second sphere at zero potential as is the case in a sphere-sphere gap with one sphere grounded. Figure 4.5 shows the two spheres separated by the centre-centre distance S.
The sphere 1 at left has potential V and that at right zero potential. Both spheres have equal radii R.
In order to hold sphere 1 at potential V, a charge Q1 = 4πe0VR must be placed at its centre. In the field of this charge, sphere 2 at right can be made a zero potential surface if an image charge Q2 is placed inside this sphere. From the discussion presented earlier,
θ1 = 10 Cµ Q2
R 0.5 m
0.125 m S
2 m
Q2 = – Q1R/S and S2 = R2/S, as shown in Figure 4.5. However, locating Q2 will disturb the potential of sphere 1. In order to keep the potential of sphere 1 undisturbed, we must locate an image charge Q1' =−Q2R/(S−S2) inside sphere 1 so that in the field of Q2 and Q'1 the potential
Fig. 4.5. Location of successive image charges to maintain spheres at potentials V and zero.
Successive image charges will have to be suitably located inside both spheres.
The sequence of charges and their locations can be tracked in a tabular form.
Example 4.5. A sphere gap consists of two spheres with R = 0.25 m each. The gap between their surfaces is 0.5 m. Calculate the charges and their locations to make the potentials 1 and 0.
Solution. The distance S between centres of the spheres = 1 m.
R2/S = 0.252/1 = 0.0625 m
Charges inside sphere 1 Charges inside sphere 2
V = 1 V = 0
Magnitude Distance from Magnitude Distance from
centre of centre of
sphere 1 sphere 2
Q1 = πe0 S1 = 0 Q2 = – 0.25 Q1 S2 = 0.0625 m
Note that further calculations will yield extremely small values for the image charges.
Furthermore, they are all located almost at the same points. The charges reduce successively in the ratio 0.25/0.933 = 0.268; i.e. Q2n =−0.268Q1nandQ1n+1 =−0.268Q2n. The electric field at any point X along the line joining the centres of the two spheres is now found from the expression
E = + charge placed at X will be in the same direction due to all charges. The most important value of X is X = R on the surface of sphere 1. If the value of E at X = R exceeds the critical gradient for breakdown of air (usually 30 kV/cm peak at an air density factor δ=1), the gap breakdown commences.
Example 4.6. Calculate the voltage gradient at X = 0.25 m for the sphere gap in Example 4.5.
The contribution of charges inside the grounded sphere amount to 693 100
Example 4.7. In the previous example calculate the potential difference between the spheres for E = 30 kV/cm = 3000 kV/m, peak.
Solution. V = 3000/4.673 kV = 642 kV
Thus, the 0.5 metre diameter spheres with a gap spacing of 0.5 metre experience disruption at 642 kV, peak. The breakdown voltage is much higher than this value.