10. Sphere or Cylinder in Uniform External Field

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ABSTRACT If a conducting or dielectric object is present within the electric field caused by external sources, then the induced charges on such an object adds to the existing external field. If the size of the conducting or dielectric object is small compared to the external field region, then the electric field is modified only in the vicinity of such an object. In order to get the complete solution of the electric field in such configuration, the boundary conditions imposed by the conducting or dielectric object must also be satisfied. If the object is of well-defined geometric shape such as cylinder or sphere, and if the external field is considered to be uniform in space, then the complete solution for the electric field could be obtained analytically with the help of method of separation of variables. These analytical solutions are often used to validate numerical methodology and could also be used to estimate field modification due to small conducting or dielectric pieces present in large high-voltage insulation arrangement.

10.1 Introduction

Conducting and dielectric components are integral parts of any electrical equipment. If the size of the conducting or dielectric object is very small compared to dimensions of the field region where the object is located, then the object contributes to the field only in the domain near the object. In many cases, such objects are present as stray bodies in high-voltage insulation arrangement. As practical examples, one may cite a small piece of conductor or dielectric floating in liquid insulation of large volume in transformers, metallic dust particles floating in gaseous insulation within gas-insulated system (GIS) and so on. It is important to understand how the presence of a conducting or dielectric object modifies the external field in the vicinity of the object, because any enhancement of the electric field intensity due to the conducting or dielectric object may lead to unwanted discharge or in the worst case failure of the insulation system.

If it is assumed that the source charges (in practical arrangement, the electrodes or conductors with specific potentials) that produce the external field is located far away from the object under consideration, then they are unaffected by the presence of the object. Consequently, the field due to the source charges may be considered to be uniform at the location of the object. If the object is such that its shape is defined by well-known mathematical functions, for example, cylinders or spheres, then the complete solution for the electric field due to the source charges located at far away positions and the induced charges on the surface of the object could be obtained by solving Laplace’s equation considering the field region to be free from any volume charge. However, in order to get the complete solution appropriate boundary conditions on the surface of the object, whether it is conducting or dielectric, need to be satisfied. One of the common methods of getting the analytical solution for a cylinder or sphere in a uniform external field is the method of separation of variables as described in this chapter.

10.2 Sphere in Uniform External Field

Consider a spherical object of radius a within uniform external field, as shown in Figure 10.1. Because the boundary is a sphere of r = constant, hence the system is best described in spherical coordinates, as shown in Figure 10.1. The uniform external field is given by ${\vec{E}}_{0} = - E_{0} {\hat{u}}_{z}$ and the potential at any point due to the external field is given by E₀r cosθ = E_0z with respect to the centre of the sphere. In order to get the complete solution for the electric field in this system, Laplace’s equation in spherical coordinates, as given in Equation 10.1, needs to be solved.

Images

FIGURE 10.1
Sphere in uniform external field.

$\frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial V}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial V}{\partial θ}) + \frac{l}{r^{2} \sin^{2} θ} \frac{\partial^{2} V}{\partial ϕ^{2}} = 0 (10.1)$

The field system has azimuthal symmetry with respect to the z-axis, that is, the field system does not change with the rotation around the z-axis. Therefore, z-axis is made the polar axis in the spherical coordinate system. Then the field is independent of coordinate φ and Laplace’s equation reduces to

$\frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial V}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial V}{\partial θ}) = 0 (10.2)$

In order to separate the terms of the left-hand side (LHS) of Equation 10.2 into functions of only one variable, Equation 10.2 may be rewritten by multiplying r² as

$\frac{\partial}{\partial r} (r^{2} \frac{\partial V}{\partial r}) + \frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial V}{\partial θ}) = 0 (10.3)$

Then LHS of Equation 10.3 is the sum of two terms, which are functions of only one variable each, that is, the first term is function of r only, whereas the second term is function of θ only. The solution to Equation 10.3 can be obtained as the product of two functions of which one is dependent only on r and the other is dependent only on θ.

Let the assumed solution be

$V (r, θ) = M (r) N (θ) (10.4)$

The assumed solution is convenient as the boundary lies at r = constant.

Combining Equations 10.3 and 10.4

$\begin{matrix} \frac{\partial}{\partial r} [r^{2} \frac{\partial M (r) N (θ)}{\partial r}] + \frac{1}{\sin θ} \frac{\partial}{\partial θ} [\sin θ \frac{\partial M (r) N (θ)}{\partial θ}] = 0 \\ or, N (θ) \frac{\partial}{\partial r} [r^{2} \frac{\partial M (r)}{\partial r}] + M (r) \frac{1}{\sin θ} \frac{\partial}{\partial θ} [\sin θ \frac{\partial N (θ)}{\partial θ}] = 0 \end{matrix}$

Dividing by M(r)N(θ),

$\frac{1}{M (r)} \frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] + \frac{1}{N (θ)} \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = 0 (10.5)$

The partial derivatives become total derivatives in Equation 10.5, as each term is dependent on only one coordinate.

The sum of two terms of the LHS of Equation 10.5 could be zero only when the two terms are separately equal to opposite and equal constant terms as given in Equation 10.6.

Equal and opposite separation constant solution:

$\frac{1}{M (r)} \frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] = + p and \frac{1}{N (θ)} \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = - p (10.6)$

where:

p is a positive constant

Another solution is obtained when the separation constant is zero.

Hence, the zero separation constant solution

$\frac{1}{M (r)} \frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] = 0 and \frac{1}{N (θ)} \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = 0 (10.7)$

Each of the above-mentioned two solutions is to be obtained separately.

Determination of the zero separation constant solution:

The first term of Equation 10.7 is

$\frac{1}{M (r)} \frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] = 0$

where:

M(r) is non-zero

Therefore,

$\frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] = 0, or, r^{2} \frac{d M (r)}{d r} = C_{0}, or, \frac{d M (r)}{d r} = \frac{C_{0}}{r^{2}}$

Integrating and incorporating constants of integration

$M (r) = \frac{C_{10}}{r} + C_{20} (10.8)$

Next the second term of Equation 10.7 is

$\frac{1}{N (θ)} \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = 0$

where:

N(θ) is non-zero

Therefore,

$\begin{matrix} \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = 0, or, \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = 0 \\ or, \sin θ \frac{d N (θ)}{d θ} = A_{0} \end{matrix}$

Integrating and incorporating constant of integration

$N (θ) = A_{10} \ln (\tan \frac{θ}{2}) + A_{20} (10.9)$

Equation 10.9 becomes undefined for θ = π. But this is not feasible in the given system as potential must be a continuous function. Therefore, A₁₀ should be zero in Equation 10.9. Therefore,

$N (θ) = A_{20} (10.10)$

Then from Equations 10.4, 10.8 and 10.10, the zero separation constant solution can be obtained as

$V (r, θ) = \frac{C_{1}}{r} + C_{2} (10.11)$

where:

C₁ = A₂₀C₁₀ and C₂ = A₂₀C₂₀

Determination of the equal and opposite separation constant solution:

The first term of Equation 10.6 is

$\begin{matrix} \frac{1}{M (r)} \frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] = + p \\ or, \frac{d}{d r} [r^{2} \frac{d M (r)}{d r}] = + p M (r) \end{matrix} (10.12)$

Putting M(r) = Crⁿ in Equation 10.12, we get

$\frac{d}{d r} (r^{2} C n r^{n - 1}) = + p C r^{n}, or, C n (n + 1) r^{n} = p C r^{n}, or, n^{2} + n - p = 0$

Hence,

$n = \frac{1}{2} (- 1 \pm \sqrt{1 + 4 p}) (10.13)$

The second term of Equation 10.6 is

$\begin{matrix} \frac{1}{N (θ)} \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = - p \\ \frac{1}{\sin θ} \frac{d}{d θ} [\sin θ \frac{d N (θ)}{d θ}] = - p N (θ) \end{matrix} (10.14)$

Putting N(θ) = Bcosθ in Equation 10.14,

$\frac{d}{d θ} (- B \sin^{2} θ) = - p B \cos θ \sin θ, or, p = 2.$

Hence, from Equation 10.13 n = +1, −2.

$Therefore, M (r) = C' r + \frac{C "}{r^{- 2}} and N (θ) = B \cos θ (10.15)$

From Equations 10.4 and 10.15,

$V (r, θ) = (C_{3} r + \frac{C_{4}}{r^{2}}) \cos θ (10.16)$

where:

C₃ = _C′_B and C₄ = C′B

The complete solution for potential function is uniquely given as a linear combination of the two solutions given by Equations 10.11 and 10.16.

$V (r, θ) = \frac{C_{1}}{r} + C_{2} + (C_{3} r + \frac{C_{4}}{r^{2}}) \cos θ (10.17)$

where:

The constants are determined by satisfying the boundary conditions

It is evident from Equation 10.17 that the first term corresponds to a net charge on the sphere and the second term to a finite potential.

10.2.1 Conducting Sphere in Uniform Field

Consider that the sphere is a conducting one and is isolated and uncharged. Further, consider that the potential at the location of the centre of the sphere due to the external field is V₀.

Because the perturbing action of the sphere is negligible at a large distance from the sphere, the potential at a large distance from the sphere (r ≫ a) is given by

$V (r, θ) = V_{0} + E_{0} r \cos θ (10.18)$

If the sphere is charged with a finite amount of charge Q, then

$V (r, θ) = \frac{Q}{4 π ε_{0} r} + V_{0} + E_{0} r \cos θ (10.19)$

In practical systems, floating metallic particles are usually not charged and hence Equation 10.18 is taken here for further discussion.

Comparing Equations 10.17 and 10.18 for r → ∞, C₂ = V₀ and C₃ = E₀. C₁ will be zero for uncharged sphere.

Therefore, Equation 10.17 can be rewritten as

$V (r, θ) = V_{0} + (E_{0} r + \frac{C_{4}}{r^{2}}) \cos θ (10.20)$

On the conductor surface, that is, for r = a,

$V (a, θ) = V_{0} + (E_{0} a + \frac{C_{4}}{a^{2}}) \cos θ (10.21)$

But the conducting sphere surface is an equipotential and hence electric potential is independent of θ on the conductor surface.

Therefore, from Equation 10.21,

C₄ = −E₀a³

Hence, the complete solution for an electric potential in the domain r > a is given by

$V (r, θ) = V_{0} + E_{0} (r - \frac{a^{3}}{r^{2}}) \cos θ (10.22)$

The r and θ components of the electric field intensity could be obtained as follows:

$E_{r} = - \frac{\partial V}{\partial r} = - E_{0} (1 + \frac{2 a^{3}}{r^{3}}) \cos θ (10.23)$

$E_{θ} = - \frac{1}{r} \frac{\partial V}{\partial θ} = E_{0} (1 - \frac{a^{3}}{r^{3}}) \sin θ (10.24)$

On the conducting sphere surface, tangential component of the electric field intensity must be zero as it is an equipotential surface. Equation 10.24 shows that for r = a, E_θ is zero, which in turn validates the solution obtained.

Again, on the conducting sphere surface, E_r is the normal component of the electric field intensity, which is given by E_r|_r=a = − 3 E₀ cos θ. Thus, the maximum value of the electric field intensity on the surface of the conducting sphere is 3E₀, that is, three times the strength of the uniform external field.

This is the reason why metallic dust particles should be avoided at all costs for GISs. Because the presence of metallic dust particles will increase the local electric field intensity three times, which will result into partial discharge within the GIS, that is very detrimental for the GIS operation.

Induced surface charge density on the surface of the conducting sphere may be obtained as follows:

$\frac{σ_{s}}{ε_{0}} = E_{r} |_{r = a} = - 3 E_{0} \cos θ, or, σ_{s} = - 3 ε_{0} E_{0} \cos θ (10.25)$

As stated earlier, the sphere may be charged with an additional charge Q, which is distributed uniformly on the sphere surface and its effect on the field could be found by superposition.

10.2.2 Dielectric Sphere in Uniform Field

In the case of dielectric sphere present in uniform external field, there will be two solutions to potential function, V_i valid for the region within the sphere having dielectric of permittivity ε_i and V_e valid for the region outside the sphere having dielectric of permittivity ε_e. Therefore, from Equation 10.17

$V_{i} (r, θ) = \frac{C_{1 i}}{r} + C_{2 i} + (C_{3 i} r + \frac{C_{4 i}}{r^{2}}) \cos θ (10.26)$

and

$V_{e} (r, θ) = \frac{C_{1 e}}{r} + C_{2 e} + (C_{3 e} r + \frac{C_{4 e}}{r^{2}}) \cos θ (10.27)$

The potential at large distance r (r ≫ a) from the sphere

$V (r, θ) = V_{0} + E_{0} r \cos θ (10.28)$

where:

V₀ is the potential at the location of the centre of the sphere due to the external field

Comparing Equations 10.27 and 10.28 for r → ∞, C_2e = V₀ and C_3e = E₀. C_1e will be zero as a dielectric sphere is not considered to have any free charge.

Hence, Equation 10.27 can be rewritten as

$V_{e} (r, θ) = V_{0} + (E_{0} r + \frac{C_{4 e}}{r^{2}}) \cos θ (10.29)$

Inside the dielectric sphere electric potential must be finite at all the points. Hence, from Equation 10.26 C_li = C_4i = 0, C_2i = V₀. Hence, Equation 10.26 can be rewritten as

$V_{i} (r, θ) = V_{0} + C_{3 i} r \cos θ (10.30)$

At r = a, both Equations 10.29 and 10.30 should yield the same electric potential. Therefore,

$\begin{matrix} V_{0} + (E_{0} a + \frac{C_{4 e}}{a^{2}}) \cos θ = V_{0} + C_{3 i} a \cos θ \\ or, E_{0} a + \frac{C_{4 e}}{a^{2}} = C_{3 i} a \end{matrix} (10.31)$

On the dielectric–dielectric boundary, the normal component of the electric field intensity should be same on both sides of the boundary. For the spherical boundary, r-component of the electric field intensity is the normal component on the boundary. Hence,

$\begin{matrix} - ε_{i} (\frac{\partial V_{i}}{\partial r}) |_{r = a} = - ε_{e} (\frac{\partial V_{e}}{\partial r}) |_{r = a} \\ or, ε_{i} C_{3 i} = ε_{e} (E_{0} - \frac{2 C_{4 e}}{a^{3}}) \end{matrix} (10.32)$

From Equations 10.31 and 10.32

$C_{4 e} = \frac{ε_{e} - ε_{i}}{2 ε_{e} + ε_{i}} a^{3} E_{0}$

and

$C_{3 i} = \frac{3 ε_{e}}{2 ε_{e} + ε_{i}} E_{0}$

Therefore, the complete solutions for potential functions inside and outside the dielectric sphere are given by

$V_{i} (r, θ) = V_{0} + \frac{3 ε_{e}}{2 ε_{e} + ε_{i}} E_{0} r \cos θ (10.33)$

and

$V_{e} (r, θ) = V_{0} + (r + \frac{ε_{e} - ε_{i}}{2 ε_{e} + ε_{i}} \frac{a^{3}}{r^{2}}) E_{0} \cos θ (10.34)$

Noting that rcos θ = z, potential function inside the dielectric sphere can be written as

$V_{i} (x, y, z) = V_{0} + \frac{3 ε_{e}}{2 ε_{e} + ε_{i}} E_{0} z (10.35)$

Hence, the electric potential within the dielectric sphere varies in only z-direction, that is, the direction of the external field. Electric field intensity within the dielectric sphere will, therefore, have only the z-component, which is given by

$E_{z i} = - \frac{\partial V_{i}}{\partial z} = - \frac{3 ε_{e}}{2 ε_{e} + ε_{i}} E_{0} (10.36)$

Images

FIGURE 10.2
Electric field in and around dielectric sphere in uniform field.

Equation 10.36 shows that the magnitude of the electric field intensity within the dielectric sphere is constant. Typical field distribution in and around a dielectric sphere within uniform external field is shown in Figure 10.2.

Equation 10.36 also shows that if ε_i < ε_e, then |E_zi| > E₀. Consider the case of a spherical air bubble trapped within a moulded solid insulation of relative permittivity 4. If the magnitude of the electric field intensity in solid insulation at the location of the air bubble is E₀, then the magnitude of the electric field intensity within the air bubble will be 1.33E₀. The operating electric field intensity within solid insulation is usually kept at a higher value as the solid insulation has a higher dielectric strength and hence such increase in field intensity within the air bubble often causes partial discharge within the air bubble as the dielectric strength of air is much lower than solid insulation.

10.3 Cylinder in Uniform External Field

Consider a long cylindrical object of radius a within uniform external field, as shown in Figure 10.3. Because the boundary is a circle of r = constant on the x–y plane, the system is best described in cylindrical coordinates, as shown in Figure 10.3. The uniform external field is given by ${\vec{E}}_{0} = - E_{0} \hat{i}$ and the potential at any point due to the external field is given by E₀r cosθ = E₀x with respect to the axis of the cylinder. In order to get the complete solution for the electric field in this system, Laplace’s equation in cylindrical coordinates as given in Equation 10.37 needs to be solved.

Images

FIGURE 10.3
Cylinder in uniform external field.

$\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial V}{\partial r}) + \frac{1}{r^{2}} \frac{\partial^{2} V}{\partial θ^{2}} + \frac{\partial^{2} V}{\partial z^{2}} = 0 (10.37)$

For this arrangement, the electric field distribution does not vary along the length of the cylinder, that is, along z-coordinate. Hence, Laplace’s equation reduces to

$\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial V}{\partial r}) + \frac{1}{r^{2}} \frac{\partial^{2} V}{\partial θ^{2}} = 0 (10.38)$

Separating the terms of the LHS into functions of only one variable by multiplying r² with Equation 10.38, it may be written that

$r \frac{\partial}{\partial r} (r \frac{\partial V}{\partial r}) + \frac{\partial^{2} V}{\partial θ^{2}} = 0 (10.39)$

The two terms on the LHS of Equation 10.39 are functions of only one variable each, that is, the first term is function of r only, whereas the second term is function of θ only. The solution to Equation 10.39 can be obtained as the product of two functions of which one is dependent only on r and the other is dependent only on θ.

Let the assumed solution be

$V (r, θ) = M (r) N (θ) (10.40)$

The assumed solution is convenient as the boundary lies at r = constant.

Combining Equations 10.39 and 10.40,

$\begin{matrix} r \frac{\partial}{\partial r} [r \frac{\partial M (r) N (θ)}{\partial r}] + \frac{\partial^{2} M (r) N (θ)}{\partial θ^{2}} = 0 \\ N (θ) {r \frac{\partial}{\partial r} [r \frac{\partial M (r)}{\partial r}]} + M (r) \frac{\partial^{2} M (r) N (θ)}{\partial θ^{2}} = 0 \end{matrix}$

Dividing by M(r)N(θ),

$\frac{r}{M (r)} \frac{d}{d r} [r \frac{d M (r)}{d r}] + \frac{1}{N (θ)} \frac{d^{2} N (θ)}{d θ^{2}} = 0 (10.41)$

The partial derivatives become total derivatives in Equation 10.41 as each term is dependent on only one coordinate.

As in the case of sphere in uniform field, zero separation constant solution and equal and opposite separation constant solution are to be obtained separately in this case, too.

Determination of the zero separation constant solution:

The first term of Equation 10.41 is

$\frac{r}{M (r)} \frac{d}{d r} [r \frac{d M (r)}{d r}] = 0$

where:

M(r) is non-zero

Therefore,

$\frac{d}{d r} [r \frac{d M (r)}{d r}] = 0, or, r \frac{d M (r)}{d r} = C_{0}, or, \frac{d M (r)}{d r} = \frac{C_{0}}{r}$

Integrating and incorporating constants of integration

$M (r) = C_{10} \ln r + C_{20} (10.42)$

Next the second term of Equation 10.41 is

$\frac{1}{N (θ)} \frac{d^{2} N (θ)}{d θ^{2}} = 0$

where:

N(θ) is non-zero

Therefore,

$N (θ) = A_{10} θ + A_{20} (10.43)$

But, from Equations 10.42 and 10.43, it can be seen that there is discontinuity of potential at r = 0 and θ = ∞, which are not feasible in the given arrangement as potential must be a continuous function. Hence, C₁₀ = A₁₀ = 0 in Equations 10.42 and 10.43.

Therefore,

$V (r, θ) = C_{20} A_{20} = C_{1} (10.44)$

Determination of the equal and opposite separation constant solution:

The first term of Equation 10.41 is

$\frac{r}{M (r)} \frac{d}{d r} [r \frac{d M (r)}{d r}] = + p$

where:

p is a positive constant

$or, r^{2} \frac{d^{2} M (r)}{d r^{2}} + r \frac{d M (r)}{d r} = + p M (r) (10.45)$

Substituting M(r) = Crⁿ in Equation 10.45, it may be obtained that

$n (n - 1) + n = p, or, n = \pm \sqrt{p}$

Hence,

$M (r) = \frac{C'}{r^{\sqrt{p}}} + C " r^{\sqrt{p}} (10.46)$

Again, the second term of Equation 10.41 is

$\begin{matrix} \frac{1}{N (θ)} \frac{d^{2} N (θ)}{d θ^{2}} = - p \\ or, \frac{d^{2} N (θ)}{d θ^{2}} = - p N (θ) \end{matrix} (10.47)$

Substituting N(θ) = e^aθ in Equation 10.47, it may be obtained that

$a^{2} e^{a θ} = - p e^{a θ}, or, a = \pm i \sqrt{p}$

Hence,

$N (θ) = B \cos (\sqrt{p} θ + α) (10.48)$

Equations 10.46 and 10.48 lead to

$V (r, θ) = M (r) N (θ) = (\frac{C_{2}}{r^{\sqrt{p}}} + C_{3} r^{\sqrt{p}}) \cos (\sqrt{p} θ + α) (10.49)$

where:

C₂ = C′B and C₃ = C″B

From Equations 10.44 and 10.49, the complete solution for potential function at all values of r and θ can be obtained as

$V (r, θ) = C_{1} + (\frac{C_{2}}{r^{\sqrt{p}}} + C_{3} r^{\sqrt{p}}) \cos (\sqrt{p} θ + α) (10.50)$

The potential at large distance r (r ≫ a) from the cylinder is given by

$V (r, θ) = V_{0} + E_{0} r \cos θ (10.51)$

Matching Equations 10.50 and 10.51, $\sqrt{p} = 1$ and α = 0.

Hence, the complete solution as given by Equation 10.50 reduces to

$V (r, θ) = C_{1} + (\frac{C_{2}}{r} + C_{3} r) \cos θ (10.52)$

10.3.1 Conducting Cylinder in Uniform Field

Comparing Equations 10.51 and 10.52, C₁ = V₀ and C₃ = E₀.

Therfore,

$V (r, θ) = V_{0} + (\frac{C_{2}}{r} + E_{0} r) \cos θ$

On the conductor surface, that is, for r = a,

$V (a, θ) = V_{0} + (\frac{C_{2}}{a} + E_{0} a) \cos θ (10.53)$

But the conducting cylinder surface is an equipotential and hence electric potential is independent of θ on the conductor surface.

Therefore, from Equation 10.53,

C₂ = −E₀a²

Hence, the complete solution for electric potential in the domain r > a is given by

$V (r, θ) = V_{0} + E_{0} (r - \frac{a^{2}}{r}) \cos θ (10.54)$

The r and θ components of the electric field intensity could be obtained as follows:

$E_{r} = - \frac{\partial V}{\partial r} = - E_{0} (1 + \frac{a^{2}}{r^{2}}) \cos θ (10.55)$

$E_{θ} = - \frac{1}{r} \frac{\partial V}{\partial θ} = E_{0} (1 - \frac{a^{2}}{r^{2}}) \sin θ (10.56)$

Equation 10.56 shows that for r = a, E_θ is zero, that is, the tangential component of the electric field intensity is zero on the cylindrical conductor surface as it is an equipotential surface.

Again, on the conducting cylinder surface, E_r is the normal component of the electric field intensity, which is given by E_r|_r=n = −2E₀ cosθ. Thus, the maximum value of the electric field intensity on the surface of the conducting cylinder is 2E₀, that is, twice the magnitude of the uniform external field. Comparing this maximum electric field intensity with the value obtained for conducting sphere in uniform field, it may be seen that the enhancement of field intensity is more if the conducting object is spherical is shape.

Induced surface charge density on the surface of the conducting cylinder may be obtained as follows:

$\frac{σ_{s}}{ε_{0}} = E_{r} |_{r = a} = - 2 E_{0} \cos θ, or, σ_{s} = - 2 ε_{0} E_{0} \cos θ (10.57)$

10.3.2 Dielectric Cylinder in Uniform Field

Potential function valid for the region within the cylinder having dielectric of permittivity ε_i is

$V_{i} (r, θ) = C_{1 i} + (\frac{C_{2 i}}{r} + C_{3 i} r) \cos θ (10.58)$

and the potential function valid for the region outside the cylinder having dielectric of permittivity ε_e is

$V_{e} (r, θ) = C_{1 e} + (\frac{C_{2 e}}{r} + C_{3 e} r) \cos θ (10.59)$

The potential at large distance r (r ≫ a) from the cylinder

$V (r, θ) = V_{0} + E_{0} r \cos θ (10.60)$

where:

V₀ is the potential at the location of the axis of the cylinder due to the external field

Comparing Equations 10.59 and 10.60 for r → ∞, C_1e = and C_3e = E₀

Hence, Equation 10.59 can be rewritten as

$V_{e} (r, θ) = V_{0} + (\frac{C_{2 e}}{r} + E_{0} r) \cos θ (10.61)$

Inside the dielectric cylinder electric potential must be finite at all the points. Hence, from Equation 10.58, C_li = V₀ and C_2i = 0. Hence, Equation 10.58 can be rewritten as

$V_{i} (r, θ) = V_{0} + C_{3 i} r \cos θ (10.62)$

At any point on the dielectric cylinder surface, that is, for r = a, electric potential as may be obtained from Equations 10.61 and 10.62 must be unique. Hence,

$\frac{C_{2 e}}{a} + E_{0} a = C_{3 i} a (10.63)$

From the boundary condition of normal component of electric flux density at r = a,

$\begin{matrix} - ε_{i} (\frac{\partial V_{i}}{\partial r}) |_{r = a} = - ε_{e} (\frac{\partial V_{e}}{\partial r}) |_{r = a} \\ or, ε_{i} C_{3 i} = ε_{e} (- \frac{C_{2 e}}{a^{2}} + E_{0}) \end{matrix} (10.64)$

From Equations 10.63 and 10.64

$\begin{matrix} C_{2 e} = \frac{ε_{e} - ε_{i}}{ε_{e} + ε_{i}} a^{2} E_{0} \\ and C_{3 i} = \frac{2 ε_{e}}{ε_{e} + ε_{i}} E_{0} \end{matrix}$

Therefore, the complete solutions for potential functions inside and outside the dielectric cylinder are given by

$V_{i} (r, θ) = V_{0} + \frac{2 ε_{e}}{ε_{e} + ε_{i}} E_{0} r \cos θ (10.65)$

$and V_{e} (r, θ) = V_{0} + (r + \frac{ε_{e} - ε_{i}}{ε_{e} + ε_{i}} \frac{a^{2}}{r}) E_{0} \cos θ (10.66)$

As rcos θ = x, potential function inside the dielectric cylinder can be written as

$V_{i} (x, y) = V_{0} + \frac{2 ε_{e}}{ε_{e} + ε_{i}} E_{0} x (10.67)$

Hence, the electric potential within the dielectric cylinder varies in only x-direction, that is, the direction of the external field. Electric field intensity within the dielectric cylinder will, therefore, have only the x-component, which is given by

$E_{x i} = - \frac{\partial V_{i}}{\partial x} = - \frac{2 ε_{e}}{ε_{e} + ε_{i}} E_{0} (10.68)$

Similar to the case of dielectric sphere in the uniform field, Equation 10.68 shows that the magnitude of the electric field intensity within the dielectric cylinder is constant. Typical field distribution on the x–y plane in and around a dielectric cylinder within uniform external field will be the same as that shown in Figure 10.2.

As in the case of dielectric sphere in uniform field, for dielectric cylinder in uniform field also |E_zi| > E₀ if ε_i < ε_e. If a cylindrical air bubble is trapped within a moulded solid insulation of relative permittivity 4, then the magnitude of the electric field intensity within the air bubble will be 1.6E₀, where E₀ is the magnitude of the electric field intensity in solid insulation at the location of the air bubble. Comparing this result with the corresponding value in the case of dielectric sphere, it may be seen that field enhancement is more if the gas cavity in liquid or solid insulation is cylindrical in shape.

Objective Type Questions

1. A conducting sphere is placed within an external field of the uniform field intensity E₀. Then the maximum value of field intensity on the surface of the conducting sphere will be

a. 0

b. E₀

c. 2E₀

d. 3E₀

2. A conducting cylinder is placed within an external field of the uniform field intensity E₀. Then the maximum value of field intensity on the surface of the conducting cylinder will be

a. 0

b. E₀

c. 2E₀

d. 3E₀

3. A dielectric sphere is placed within an external field of the uniform field intensity E₀. The external field intensity acts in the direction of the x-axis. Then the field intensity within the dielectric sphere will be in the direction of

a. x-axis

b. y-axis

c. z-axis

d. None of the above

4. A dielectric cylinder is placed within an external field of the uniform field intensity E₀. The external field intensity acts in the direction of y-axis. Then the field intensity within the dielectric cylinder will be in the direction of

a. x-axis

b. y-axis

c. z-axis

d. None of the above

5. A dielectric sphere is placed within an external field of the uniform field intensity E₀. Then the field intensity within the dielectric sphere will be

a. Zero

b. Equal to E₀

c. Constant but not equal to E₀

d. None of the above

6. A dielectric cylinder is placed within an external field of the uniform field intensity E₀. Then the field intensity within the dielectric cylinder will be

a. Zero

b. Equal to E₀

c. Constant but not equal to E₀

d. None of the above

7. A gas cavity is present in large volume of solid insulation. Then the ratio of field intensity within the gas cavity for spherical to that for cylindrical cavity will be

a. Zero

b. <1

c. 1

d. >1

8. A conducting object is present in an external field of uniform field intensity. Then the ratio of maximum field intensity on the surface of the conducting object for spherical to that for cylindrical object will be

a. Zero

b. <1

c. 1

d. >1

Answers:

1) d;

2) c;

3) a;

4) b;

5) c;

6) c;

7) b;

8) d

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Table of Contents for 10. Sphere or Cylinder in Uniform External Field

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Table of Contents for
10. Sphere or Cylinder in Uniform External Field