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Electromags - Chapter 6

If we wish to move the charge in the direction of the field, our energ...

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Electronics Engineering (CR 061)

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LESSON CONTENT

If we wish to move the charge in the direction of the field, our energy expenditure turns out to be negative; we do not do the work, the field does.

Suppose we wish to move a charge Q a distance dL in an electric field E. The force on Q arising from the electric field is

(1)

where the subscript reminds us that this force arises from the field. The component of this force in the direction dL which we must overcome is

where aL = a unit vector in the direction of dL.

The force that we must apply is equal and opposite to the force associated with the field,

and the expenditure of energy is the product of the force and distance. That is, the differential work done by an external source moving charge Q is dW = −QE · aLdL,

or (2)

where we have replaced aLdL by the simpler expression dL.

This differential amount of work required may be zero under several conditions determined easily from Eq. (2). There are the trivial conditions for which E, Q, or dL is zero, and a much more important case in which E and dL are perpendicular. Here the charge is moved always in a direction at right angles to the electric field. We can draw on a good analogy between the electric field and the gravitational field, where, again, energy must be expended to move against the field. Sliding a mass around with constant velocity on a frictionless surface is an effortless process if the mass is moved along a constant elevation contour; positive or negative work must be done in moving it to a higher or lower elevation, respectively.

Returning to the charge in the electric field, the work required to move the charge a finite distance must be determined by integrating,

(3) where the path must be specified before the integral can be evaluated. The charge is assumed to be at rest at both its initial and final positions. This definite integral is basic to field theory, and we shall devote the following section to its interpretation and evaluation.

Example:

The value of E at P(ρ = 2, φ = 40°, z = 3) is given as E = 100aρ − 200aφ + 300az V/m. Determine the incremental work required to move a 20 μC charge a distance of 6μm: (a) in the direction of aρ; (b) in the direction of aφ; (c) in the direction of az; (d) in the direction of E; (e) in the direction of G = 2ax − 3ay + 4az.

Solution:

a) in the direction of aρ: The incremental work is given by dW = −qE ・ dL, where in this case,

dL = dρaρ = 6× 10 − 6 aρ. Thus

dW = −(20 × 10 − 6 C)(100V/m)(6 × 10 − 6 m) = − 12 × 10 − 9 J = −12 nJ

b) in the direction of aφ: In this case dL = 2dφ aφ = 6× 10 −6 aφ, and so

dW = −(20 × 10 − 6 )(−200)(6 × 10 − 6 ) = 2. 4 × 10 − 8 J = 24 nJ

c) in the direction of az: Here, dL = dz az = 6× 10 − 6 az, and so

dW = −(20 × 10 − 6 )(300)(6 × 10 − 6 ) = − 3. 6 × 10 − 8 J = −36 nJ

d) in the direction of E: Here, dL = 6× 10 − 6 aE, where

Thus dW = −(20 × 10 − 6 )[100aρ − 200 aφ + 300az] ・ [0. 267 aρ − 0. 535 aφ + 0. 802 az](6 ×

10 − 6 ) = − 44 .9nJ

e) In the direction of G = 2ax − 3 ay + 4az: In this case, dL = 6× 10 − 6 aG, where

So now

dW = −(20 × 10 − 6 )[100aρ − 200 aφ + 300az] ・ [0. 371 ax − 0. 557 ay + 0. 743 az](6 × 10 −6)

= −(20 × 10 − 6 ) [37(aρ・ ax) − 55 .7(aρ・ ay) − 74 .2(aφ・ ax) + 111(aφ ・ ay)

+ 222] (6 × 10 − 6 )

where, at P, (aρ ・ ax) = (aφ ・ ay) = cos(40°) = 0, (aρ ・ ay) = sin(40°) = 0, and

(aφ ・ ax) = −sin(40°) = − 0 .643. Substituting these results in

dW = −(20 × 10 − 6 )[28. 4 − 35 .8 + 47 + 85 + 222](6 × 10 − 6 ) = − 41 .8nJ

Example: An electric field is given as E = − 10 ey(sin2z ax + xsin2z ay + 2xcos2z az) V/m. a) Find E at P(5, 0 , π/12). b) How much work is done in moving a charge of 2 nC an incremental distance of 1 mm from P in the direction of ax? c) of ay? d) of az? e) of (ax + ay + az)?

Solution:

a) Find E at P(5, 0 , π/12): Substituting this point into the given field produces

EP = −10 [sin(π/6) ax + 5sin(π/6) ay + 10cos(π/6) az] = − [5az + 25ay

b) How much work is done in moving a charge of 2 nC an incremental distance of 1 mm

because we have assumed a uniform field,

What is this sum of vector segments in the preceding parentheses? Vectors add by the parallelogram law, and the sum is just the vector directed from the initial point B to the final point A, LBA. Therefore

(4)

Remembering the summation interpretation of the line integral, this result for the uniform field can be obtained rapidly now from the integral expression

applied to a uniform field

where the last integral becomes LBA and

For this special case of a uniform electric field intensity, we should note that the work involved in moving the charge depends only on Q, E, and LBA, a vector drawn from the initial to the final point of the path chosen. It does not depend on the particular path we have selected along which to carry

and

(5) as

the charge may proceed from B to A on a straight line or via the Old Chisholm Trail; the answer is the same. We show in Section 4 that an identical statementmaybe made for any nonuniform (static)Efield. Let us use several examples to illustrate the mechanics of setting up the line integral appearing in Eq. (5).

Example: We are given the nonuniform field

E = yax + xay + 2az

and we are asked to determine the work expended in carrying 2C from B(1, 0 , 1) to A(0. 8 , 0. 6 , 1) along the shorter arc of the circle x 2

where the limits on the integrals have been chosen to agree with the initial and final values of the appropriate variable of integration. Using the equation of the circular path (and selecting the sign of the radical which is correct for the quadrant involved), we have

Example: Again find the work required to carry 2C from B to A in the same field, but this time use the straight-line path from B to A. Solution. We start by determining the equations of the straight line. Any two of the following three equations for planes passing through the line are sufficient to define the line:

y 2 = 1,z = 1

Solution: We use W = −Q 𝐀 𝐀

whereE is not necessarily constant. Working in rectangular

coordinates, the differential pathdL is dxax + dyay + dzaz , and the integral becomes

The interrelationships among the several variables in each expression are determined from the specific equations for the path.

As a final example illustrating the evaluation of the line integral, we investigate several paths that we might take near an infinite line charge. The field has been obtained several times and is entirely in the radial direction,

First we find the work done in carrying the positive charge Q about a circular path of radius ρb centered at the line charge, as illustrated in Figure 4. Without lifting a pencil, we see that the work must be nil, for the path is always perpendicular to the electric field intensity, or the force on the charge is always exerted at right angles to the direction in which we are moving it. For practice, however, we will set up the integral and obtain the answer.

The differential element dL is chosen in cylindrical coordinates, and the circular path selected demands that dρ and dz be zero, so dL = ρ 1 dØaØ. The work is then

Because b is larger than a, ln(b/a) is positive, and the work done is negative, indicating that the external source that is moving the charge receives energy.

One of the pitfalls in evaluating line integrals is a tendency to use too many minus signs when a charge is moved in the direction of a decreasing coordinate value. This is taken care of completely by the limits on the integral, and no misguided attempt should be made to change the sign of dL. Suppose we carry Q from b to a (Figure 4). We still have dL = dρaρ and show the different direction by recognizing ρ = b as the initial point and ρ = a as the final point,

This is the negative of the previous answer and is obviously correct.

Example: Let G = 3xy 3 ax + 2zay. Given an initial point P(2, 1 , 1) and a final point Q(4, 3 , 1), find 𝐀

・ 𝐀𝐀 using the path: a) straight line: y = x − 1, z = 1, b) parabola: 6y = x 2 + 2, z = 1

Solution:

a) straight line: y = x − 1, z = 1: We obtain:

parabola: 6y = x 2 + 2, z = 1: We obtain:

Definition of Potential Difference and Potential

We are now ready to define a new concept from the expression for the work done by an external source in moving a charge Q from one point to another in an electric field E, “Potential difference and work.”

In much the same way as we defined the electric field intensity as the force on a unit test charge, we now define potential difference V as the work done (by an external source) in moving a unit positive charge from one point to another in an electric field, (9)

We have to agree on the direction of movement, and we do this by stating that VAB signifies the potential difference between points A and B and is the work done in moving the unit charge from B (last named) to A (first named). Thus, in determining VAB, B is the initial point and A is the final point. The reason for this somewhat peculiar definition will become clearer shortly, when it is seen that the initial point B is often taken at infinity, whereas the final point A represents the fixed position of the charge; point A is thus inherently more significant. Potential difference is measured in joules per coulomb, for which the volt is defined as a more common unit, abbreviated as V. Hence the potential difference between points A and B is

(10)

and VAB is positive if work is done in carrying the positive charge from B to A.

From the line-charge example of Section 4 we found that the work done in taking a charge Q from ρ= b to ρ= a was

Thus, the potential difference between points at ρ = a and ρ= b is

(11)

We can try out this definition by finding the potential difference between points A and B at radial distances rA and rB from a point charge Q. Choosing an origin at Q,

a) Find the absolute potential at P(r = 1cm, θ = 25°, φ = 50°): Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter. The potential function for r > 0 .6 cm will be that of a point charge of Q = 4πa 2 ρs, or

At r = 1cm, this becomes V (r = 1cm) = 8 .14 V

b) Find VAB given points A(r = 2cm, θ = 30◦, φ = 60◦) and B(r = 3cm, θ = 45◦, φ = 90◦): Again, the angles do not matter because of the spherical symmetry. We use the part a result to obtain

Example: Let a uniform surface charge density of 5 nC/m 2 be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4, and a point charge of 2μC be present at P(2, 0 , 0). If V = 0 at M(0, 0 , 5), find V at N(1, 2 , 3).

Solution: We need to find a potential function for the combined charges which is zero at M. That for the point charge we know to be

Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have

For the sheet charge, we have

The total potential function will be the sum of the three. Combining the integration constants, we obtain:

The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r, ρ, and z are scalar distances from the charges, and will be treated as such here. To evaluate the constant, C, we first look at point M, where VT = 0.

Potential Field of a Point Charge

In Eq. (12) for the potential difference between two points located at r = rA and r = rB in the field of a point charge Q placed at the origin. How might we conveniently define a zero reference for potential? The simplest possibility is to let V = 0 at infinity. If we let the point at r = rB recede to infinity, the potential at rA becomes

or, as there is no reason to identify this point with the A subscript,

(14)

This expression defines the potential at any point distant r from a point charge Q at the origin, the potential at infinite radius being taken as the zero reference. Returning to a physical interpretation, we may say that Q/ 4 𝐀𝐀 0 r joules of work must be done in carrying a unit charge from infinity to any point r meters from the charge Q.

A convenient method to express the potential without selecting a specific zero reference entails identifying rA as r once again and letting Q/ 4 𝐀𝐀 0 rB be a constant. Then

(15)

and C 1 may be selected so that V = 0 at any desired value of r. We could also select the zero reference indirectly by electing to let V be V 0 at r = r 0.

It should be noted that the potential difference between two points is not a function of C 1.

Equations (14) and (15) represent the potential field of a point charge. The potential is a scalar field and does not involve any unit vectors. We now define an equipotential surface as a surface composed of all those points having the same value of potential. All field lines would be perpendicular to such a surface at the points where they intersect it. Therefore, no work is involved in moving a unit charge around on an equipotential surface. The equipotential surfaces in the potential field of a point charge are spheres centered at the point charge.

An inspection of the form of the potential field of a point charge shows that it is an inversedistance field, whereas the electric field intensity was found to be an inverse-square-law function. A similar result occurs for the gravitational force field of a point mass (inverse-square law) and the gravitational potential field (inverse distance). The gravitational force exerted by the earth on an object one million miles from it is four times that exerted on the same object two million miles away. The kinetic energy given to a freely falling object starting from the end of the universe with zero velocity, however, is only twice as much at one million miles as it is at two million miles.

Example: In spherical coordinates, E = 2r/(r 2 + a 2 ) 2 ar V/m. Find the potential at any point, using the reference, a) V = 0 at infinity, b) V = 0 at r = 0, c) V = 100V at r = a Solution: a) V = 0 at infinity: We write in general

b) V = 0 at r = 0: Using the general expression, we find

We have come quite a distance from the potential field of the single point charge, and it might be helpful to examine Eq. (17) and refresh ourselves as to the meaning of each term. The potential V(r) is determined with respect to a zero reference potential at infinity and is an exact measure of the work done in bringing a unit charge from infinity to the field point at r where we are finding the potential. The volume charge density ρv(r’) and differential volume element dv’ combine to represent a differential amount of charge ρv(r’)dv’ located at r’. The distance |r – r’| is that distance from the source point to the field point. The integral is a multiple (volume) integral.

If the charge distribution takes the form of a line charge or a surface charge, the integration is along the line or over the surface:

The most general expression for potential is obtained by combining Eqs.(16)–(19). These integral expressions for potential in terms of the charge distribution should be compared with similar expressions for the electric field intensity, such as Eq. (15) in Field Due to Continuous Charge Distribution:

(18)

The potential again is inverse distance, and the electric field intensity, inverse square law. The latter, of course, is also a vector field.

Example: To illustrate the use of one of these potential integrals, we will find V on the z axis for uniform line charge ρL in the form of a ring, ρ = a, in the z = 0 plane, as shown in Figure 4.

Solution: Working with Eq. (18), we have dL’ = adØ’ , r = za , r’

, and For a zero reference at infinity, then: a. The potential arising from a single point charge is the work done in carrying a unit positive charge from infinity to the point at which we desire the potential, and the work is independent of the path chosen between those two points.

b. The potential field in the presence of a number of point charges is the sum of the individual potential fields arising from each charge.

c. The potential arising from a number of point charges or any continuous charge distribution may therefore be found by carrying a unit charge from infinity to the point in question along any path we choose.

or potential difference,

regardless of the source of the E field.

This result is often stated concisely by recognizing that no work is done in carrying the unit charge around any closed path, or

A small circle is placed on the integral sign to indicate the closed nature of the path. This symbol also appeared in the formulation of Gauss’s law, where a closed surface integral was used Equation (20) is true for static fields, but Faraday demonstrated it was incomplete when timevarying magnetic fields were present. One of Maxwell’s greatest contributions to electromagnetic theory was in showing that a time-varying electric field produces a magnetic field, and therefore we should expect to find later that Eq. (20) is not correct when either E or the magnetic field varies with time.

Restricting our attention to the static case where E does not change with time, consider the dc circuit shown in Figure 4. Two points, A and B, are marked, and (20) states that no work is involved in carrying a unit charge from A through R 2 and R 3 to B and back to A through R 1 , or that the sum of the potential differences around any closed path is zero.

composed of wires, resistances, and batteries. Equation (20) must be amended before we can apply it to time-varying fields.

Any field that satisfies an equation of the form of Eq. (20), (i., where the closed line integral of the field is zero) is said to be a conservative field. The name arises from the fact that no work is done (or that energy is conserved) around a closed path. The gravitational field is also conservative, for any energy expended in moving (raising) an object against the field is recovered exactly when the object is returned (lowered) to its original position. A nonconservative gravitational field could solve our energy problems forever.

In other words, the expression for potential (zero reference at infinity),

is not dependent on the path chosen for the line integral,

Equation (20) is therefore just a more general form of Kirchhoff’s circuital law for voltages, more general in that we can apply it to any region where an electric field exists and we are not restricted to a conventional circuit

Now consider a general region of space, as shown in Figure 4, in which E and V both change as we move from point to point. Equation (22) tells us to choose an incremental vector element of length ΔL = ΔLaL and multiply its magnitude by

the component of E in the direction of aL (one interpretation of the dot product) to obtain the small potential difference between the final and initial points of ΔL.

If we designate the angle between ΔL and E as Θ, then

We now pass to the limit and consider the derivative dV/dL. To do this, we need to show that V may be interpreted as a function V(x, y, z). So far, V is merely the result of the line integral (21). If we assume a specified starting point or zero reference and then let our end point be (x, y, z), we know that the result of the integration is a unique function of the end point (x, y, z) because E is a conservative field. Therefore V is a single-valued function V(x, y, z). We may then pass to the limit and obtain

In which direction should ΔL be placed to obtain a maximum value of ΔV? Remember that E is a definite value at the point at which we are working and is independent of the direction of ΔL. The magnitude ΔL is also constant, and our variable is aL , the unit vector showing the direction of ΔL. It is obvious that the maximum positive increment of potential, ΔVmax, will occur when cosΘ is −1, or ΔL points in the direction opposite to E. For this condition,

This little exercise shows us two characteristics of the relationship between E and V at any point: a. The magnitude of the electric field intensity is given by the maximum value of the rate of change of potential with distance. b. This maximum value is obtained when the direction of the distance increment is opposite to E or, in other words, the direction of E is opposite to the direction in which the potential is increasing the most rapidly.

We now illustrate these relationships in terms of potential. Figure 4 is intended to show the information we have been given about some potential field. It does this by showing the equipotential

surfaces (shown as lines in the two-dimensional sketch). We desire information about the electric field intensity at point P. Starting at P, we lay off a small incremental distance ΔL in various directions, hunting for that direction in which the potential is changing (increasing) the most rapidly. From the sketch, this direction appears to be left and slightly upward. From our second characteristic above, the electric field intensity is therefore oppositely directed, or to the right and slightly downward at P. Its magnitude is given by dividing the small increase in potential by the small element of length.

It seems likely that the direction in which the potential is increasing the most rapidly is perpendicular to the equipotentials (in the direction of increasing potential), and this is correct, for if ΔL is directed along an equipotential, ΔV = 0 by our definition of an equipotential surface. But then

and as neither E nor ΔL is zero, E must be perpendicular to this ΔL or perpendicular to the equipotentials.

Because the potential field information is more likely to be determined first, let us describe the direction of ΔL, which leads to a maximum increase in potential mathematically in terms of the potential field rather than the electric field intensity. We do this by letting aN be a unit vector normal to the equipotential surface and directed toward the higher potentials. The electric field intensity is then expressed in terms of the potential,

(23)

which shows that the magnitude of E is given by the maximum space rate of change of V and the direction of E is normal to the equipotential surface (in the direction of decreasing potential).

Because dV/dL|max occurs when _L is in the direction of aN, we may remind ourselves of this fact by letting

and (24)

Either Eq. (23) or Eq. (24) provides a physical interpretation of the process of finding the electric field intensity from the potential. Both are descriptive of a general procedure, and we do not intend to use them directly to obtain quantitative information.

The gradient of a scalar is a vector, and old quizzes show that the unit vectors that are often incorrectly added to the divergence expression appear to be those that were incorrectly removed from the gradient. Once the physical interpretation of the gradient, expressed by Eq. (25), is grasped as showing the maximum space rate of change of a scalar quantity and the direction in which this maximum occurs, the vector nature of the gradient should be self-evident.

The vector operator

may be used formally as an operator on a scalar, T , T , producing

from which we see that

This allows us to use a very compact expression to relate E and V, (29)

The gradient may be expressed in terms of partial derivatives in other coordinate systems through the application of its definition Eq. (25).

(30)

(31)

(32)

Note that the denominator of each term has the form of one of the components of dL in that coordinate system, except that partial differentials replace ordinary differentials; for example, r sinΘ

dØ becomes r sinΘ∂Ø.

We now illustrate the gradient concept with an example.

Example: Given the potential field, V = 2x 2 y − 5 z, and a point P(− 4 , 3 , 6), we wish to find several numerical values at point P: the potential V, the electric field intensity E, the direction of E, the electric flux density D, and the volume charge density ρv. Solution: The potential at P(−4, 5, 6) is VP = 2(−4)2(3) − 5(6) = 66 V

Next, we may use the gradient operation to obtain the electric field intensity, E = − V = −4xyax − 2x 2 ay + 5az V/m

The value of E at point P is

∇

EP = 48ax − 32ay + 5az V/m and

|EP| = = 57 V/m

The direction of E at P is given by the unit vector aE,P = (48ax − 32ay + 5az )/57 = 0 − 0 + 0

If we assume these fields exist in free space, then D = 𝐀 0 E = −35 ax − 17 2 ay + 44 az pC/m 3

Finally, we may use the divergence relationship to find the volume charge density that is the source of the given potential field, ρν = ·D = −35 pC/m 3 At P, ρν = −106 pC/m 3.

Example: Let V = 2xy 2 z 3 + 3 ln(x 2 + 2y 2 + 3z 2 ) V in free space. Evaluate each of the following quantities at P(3, 2 ,−1) (a) V; (b) |V|; (c) E; (d) |E|; (e) aN ; ( f ) D.

Solution:

a) V : Substitute P directly to obtain: V = − 15 .0V

b) |V |. This will be just 15 .0V.

c) E: We have

d) |E|P : taking the magnitude of the part c result, we find |E|P = 75 .0V/m.

e) aN: By definition, this will be

f) D: This is D|P = 𝐀 0 E|P = 62 .8 ax + 202 ay −

629 az pC/m.

The Dipole

The dipole fields that we develop in this section are quite important because they form the basis for the behavior of dielectric materials in electric fields. Moreover, this development will serve to illustrate the importance of the potential concept presented in this module.

An electric dipole, or simply a dipole, is the name given to two point charges of equal magnitude and opposite sign, separated by a distance that is small compared to the distance to the point P at

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Electromags - Chapter 6

Course: Electronics Engineering (CR 061)

95 Documents

Students shared 95 documents in this course

University: Samar State University

Was this document helpful?

LESSON CONTENT

If we wish to move the charge in the direction of the field, our energy expenditure turns out to be negative; we

do not do the work, the field does.

Suppose we wish to move a charge Q a distance dL in an electric field E. The force on Q arising

from the electric field is

(1)

where the subscript reminds us that this force arises from the field. The component of this

force in the direction dL which we must overcome is

where aL = a unit vector in the direction of dL.

The force that we must apply is equal and opposite to the force associated with the field,

and the expenditure of energy is the product of the force and distance. That is, the differential work

done by an external source moving charge Q is dW = −QE · aLdL,

or (2)

where we have replaced aLdL by the simpler expression dL.

This differential amount of work required may be zero under several conditions determined

easily from Eq. (2). There are the trivial conditions for which E, Q, or dL is zero, and a much more

important case in which E and dL are perpendicular. Here the charge is moved always in a direction

at right angles to the electric field. We can draw on a good analogy between the electric field and

the gravitational field, where, again, energy must be expended to move against the field. Sliding a

mass around with constant velocity on a frictionless surface is an effortless process if the mass is

moved along a constant elevation contour; positive or negative work must be done in moving it to a

higher or lower elevation, respectively.

Returning to the charge in the electric field, the work required to move the charge a finite

distance must be determined by integrating,

(3) where

the path must be specified before the integral can be evaluated. The charge is assumed

to be at rest at both its initial and final positions. This definite integral is basic to field theory, and we shall

devote the following section to its interpretation and evaluation.

Example:

Electromags - Chapter 6

Electronics Engineering (CR 061)

Samar State University

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LESSON CONTENT

(1)

a) in the direction of aρ: The incremental work is given by dW = −qE ・ dL, where in this case,

Thus dW = −(20 × 10 − 6 )[100aρ − 200 aφ + 300az] ・ [0. 267 aρ − 0. 535 aφ + 0. 802 az](6 ×

e) In the direction of G = 2ax − 3 ay + 4az: In this case, dL = 6× 10 − 6 aG, where

dW = −(20 × 10 − 6 )[100aρ − 200 aφ + 300az] ・ [0. 371 ax − 0. 557 ay + 0. 743 az](6 × 10 −6)

= −(20 × 10 − 6 ) [37(aρ・ ax) − 55 .7(aρ・ ay) − 74 .2(aφ・ ax) + 111(aφ ・ ay)

+ 222] (6 × 10 − 6 )

where, at P, (aρ ・ ax) = (aφ ・ ay) = cos(40°) = 0, (aρ ・ ay) = sin(40°) = 0, and

(aφ ・ ax) = −sin(40°) = − 0 .643. Substituting these results in

dW = −(20 × 10 − 6 )[28. 4 − 35 .8 + 47 + 85 + 222](6 × 10 − 6 ) = − 41 .8nJ

EP = −10 [sin(π/6) ax + 5sin(π/6) ay + 10cos(π/6) az] = − [5az + 25ay

(4)

・ 𝐀𝐀 using the path: a) straight line: y = x − 1, z = 1, b) parabola: 6y = x 2 + 2, z = 1

a) straight line: y = x − 1, z = 1: We obtain:

(10)

(11)

(14)

(18)

(23)

(30)

(31)

(32)

dØ becomes r sinΘ∂Ø.

∇

a) V : Substitute P directly to obtain: V = − 15 .0V

f) D: This is D|P = 𝐀 0 E|P = 62 .8 ax + 202 ay −

Electromags - Chapter 6

Course: Electronics Engineering (CR 061)

University: Samar State University

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