Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 28

Accordingly, we have E ≈ kQ/a2, which resembles the magnitude of the electric field produced by a point charge.. 20.13 A rod of length L has a uniform positive charge densityλ and an ele

dE = k dq

x2 = k λ dx

The total electric field at P due to all the segments of the rod is given by Eq.20.20

after integrating from one end of the rod (x = a) to the other (x = a + L) as follows:

E = dE=

a+L a

k λ dx

x2 = kλ

a+L a

x−2dx = kλ−1

xa +L a

= kλ



a + L +

1

a



= k λL

a (a + L)

(20.26)

When we use the fact that the total charge is Q = λL, we have:

E= kQ

If P is a very far point from the rod, i.e a  L, then L can be neglected in the

denominator of Eq.20.27 Accordingly, we have E ≈ kQ/a2, which resembles the

magnitude of the electric field produced by a point charge

For a Point on the Perpendicular Bisector of the Rod

A rod of length L has a uniform positive charge density λ and total charge Q The rod is placed along the x-axis as shown in Fig.20.13 Assume that point P is on the perpendicular bisector of the rod and is located at a constant distance a from the origin of the x-axis The charge on a segment dx on the rod will be dq = λ dx.

Fig 20.13 A rod of length L

has a uniform positive charge

densityλ and an electric field

d→

E at point P due to a segment

of charge dq, where P is

located along the

perpendicular bisector of the

rod From symmetry, the total

field will be along the y-axis

x

0

a

L

P

dx

dq

+ + + + + + + + + + + +

θ θ ο

x

d E

x

d E

y

d E

r y

The electric field d→

E at P due to this segment has a magnitude:

dE = k dq

r2 = k λ dx

This field has a vertical component dEy = dE sin θ along the y-axis and a horizontal component dExperpendicular to it, as shown in Fig.20.13 An x-component at such

a location is canceled out by a similar but symmetric charge segment on the opposite side of the rod Thus:

The total electric field at P due to all segments of the rod is given by two times the integration of the y-component from the middle of the rod (x= 0) to one of the

ends (x = L/2) Thus:

E= 2

x =L/2

x=0

dEy= 2

x =L/2

x=0

dE sin θ = 2 k λ

x =L/2

x=0

sinθ dx

To perform the integration of this expression, we must relate the variablesθ, x, and

r One approach is to express θ and r in terms of x From the geometry of Fig.20.13,

we have:

r=x2+ a2 and sinθ = a

r = √ a

x2+ a2 (20.31) Therefore, Eq.20.30becomes:

E = 2 kλ a

L/2

0

dx (x2+ a2)3/2 (20.32)

From the table of integrals in Appendix B, we find that:

(x2+ a2)3/2 = x

a2

(x2+ a2) (20.33)

Thus:

E = 2kλ a

L/2

0

dx (x2+ a2)3/2 = 2 kλ a

a2√ x

x2+ a2



L /2

0

= 2kλ a



L /2

a2

(L/2)2+ a2− 0



a

(L/2)2+ a2 (20.34)

When we use the fact that the total charge is Q = λL, we have:

a

a2+ (L/2)2 or E = k λ L

a

a2+ (L/2)2 (20.35)

When P is a very far point from the rod, a  L, we can neglect (L/2)2in the denom-inator of Eq.20.35 Thus, E ≈ kQ/a2 This is just the form of a point charge For an

infinitely long rod we get:

E= lim

L→∞

2k λ

a

(2a/L)2+ 1 ⇒ E = 2k

λ

Example 20.4

Figure 20.14 shows a non-conducting rod that has a uniform positive charge density +λ and a total charge Q along its right half, and a uniform negative

charge density−λ and a total charge −Q along its left half What is the direction and magnitude of the net electric field at point P that shown in Fig.20.14?

Fig 20.14

x

0

L

P y

a

+Q

-Q

Solution: When we consider a segment dx on the right side of the rod, the charge

on this segment will be dq = λ dx, see Fig.20.15

The electric field d→

E+at P due to this segment is directed outwards and away from the positive charge dq and has a magnitude:

dE+= k dq

r2 = k λ dx

r2

A symmetric segment on the opposite side of the rod, but with a negative charge,

creates an electric field d→

E−that is directed inwards and toward this segment and

has the same magnitude as d→

E+, i.e dE+= dE− The resultant electric field d E→

from both symmetric segments will be a vector to the left, see Fig.20.15, and its magnitude will be given by:

dE = dE+cosθ + dE+cosθ = 2 dE+cosθ

= 2 k λ dx

r2

x

r = k λ(x2+ a2) −3/2 (2x)dx

Fig 20.15

x

0

L

P

+

y

a

+Q

- Q

θ

dq r r

d E+

d E−

θ

d E

The total electric field at P due to all segments of the rod is found by inte-grating dE from x = 0 to only x = L/2,since the negative charge of the rod is considered in evaluating dE Thus:

E= dE = k λ

x =L/2

x=0

(x2+ a2) −3/2 (2x dx)

To evaluate the integral in this equation, we transform it to the form



u n du = u n+1/(n + 1), as we shall do in solving Eq.20.53 Thus:

E = k λ

(u2+ a −1/22) −1/2u =L/2

u=0 = k λ



−2



(L/2)2+ a2 −−2

a



= 2k λ

 1

(L/2)2+ a2



When we use the fact that the magnitude of the charge Q is given by Q = λL/2,

we get:

E =4k Q

L

 1

(L/2)2+ a2



When P is very far away from the rod, i.e a  L, we can neglect (L/2)2 in

the denominator of this equation and hence get E ≈ 0 In this situation, the two oppositely charged halves of the rod would appear to point P as if they were two

coinciding point charges and hence have a zero net charge

Example 20.5

An infinite sheet of charge is lying on the xy-plane as shown in Fig.20.16 A positive charge is distributed uniformly over the plane of the sheet with a charge per unit areaσ Calculate the electric field at a point P located a distance a from

the plane

x

y

+ + +

+ +

+

a

d x x

P

z dE x

dE

d E

o

r z

θ

Fig 20.16

Solution: Let us divide the plane into narrow strips parallel to the y-axis.

A strip of width dx can be considered as an infinitely long wire of charge per

unit lengthλ = σ dx From Eq.20.36, at point P, the strip sets up an electric field

d→

E lying in the xz-plane of magnitude:

dE = 2k λ

r = 2k σdx

r

This electric field vector can be resolved into two components d→

Ex and d→

Ez

By symmetry the components d→

Exwill sum to zero when we consider the entire

sheet of charge Therefore, the resultant electric field at point P will be in the z-direction, perpendicular to the sheet From Fig.20.16, we find the following:

dEz = dE sin θ

and hence:

E= dEz = 2kσ

+∞

−∞

sinθdx r

To perform the integration of this expression, we must first relate the variablesθ,

x, and r One approach is to express θ and r in terms of x From the geometry of

Fig.20.16, we have:

r=x2+ a2 and sinθ = a

r = √ a

x2+ a2

Then, from the table of integrals in Appendix B, we find that:

E = 2kσ a

+∞

−∞

dx

x2+ a2 = 2kσ a

1atan−1x

a



+∞−∞

= 2kσtan−1(∞) − tan−1(−∞)= 2kσ π

2 +π 2

 Thus:

E = 2πkσ = σ

2◦

This result is identical to the one we shall find inSect 20.4.4for a charged

disk of infinite radius We note that the distance a from the plane to the point P does not appear in the final result of E This means that the electric field set up at

any point by an infinite plane sheet of charge is independent of how far the point

is from the plane In other words, the electric field is uniform and normal to the

plane

Also, the same result is obtained if the point P lies below the xy-plane That

is, the field below the plan has the same magnitude as that above the plane but as

a vector it points in the opposite direction

20.4.2 The Electric Field of a Uniformly Charged Arc

Assume that a rod has a uniformly distributed total positive charge Q Also assume that the rod is bent into a circular section of radius R and central angle φ rad To find the electric field at the center P of this arc, we place coordinate axes such that the axis of symmetry of the arc lies along the y-axis and the origin is at the arc’s center,

see Fig.20.17a If we letλ represent the linear charge density of this arc which has

a length R φ, then:

For an arc element ds subtending an angle d θ at P, we have:

Therefore, the charge dq on this arc element will be given by:

dq = λ ds = Q

R φ R d θ =

Q

To find the electric field at point P, we first calculate the magnitude of the electric field dE at P due to this element of charge dq, see Fig.20.17b, as follows:

dE = k dq

R2 = kQ

This field has a vertical component dEy = dE cos θ along the y-axis and a horizontal component dEx along the negative x-axis, as shown in Fig.20.17b The x-component created at P by any charge element dq is canceled by a symmetric charge element on

the opposite side of the arc Thus, the perpendicular components of all of the charge elements sum to zero The vertical component will take the form:

dEy = dE cos θ = kQ

R2φcosθ dθ (20.41)

Consequently, the total electric field at P due to all elements of the arc is given by the integration of the y-component as follows:

E= dEy= kQ

R2φ

+φ/2

−φ/2

cosθ dθ = kQ

R2φsin θ+φ/2

−φ/2= kQ

R2φ

 sinφ

2 − sin −φ

2

 (20.42)

P y

x

φ

(a)

R Q

R

P

d E

y

d E y

d q

θ

R Q

x dE

(b)

R

d s s

x

dθ

Fig 20.17 (a) A circular arc of radius R, central angle φ, and center P has a uniformly distributed positive charge Q (b) The figure shows the electric field d→E at P due to an arc element ds having a charge

dq From symmetry, the horizontal components of all elements cancel out and the total field is along the y-axis

Finally, the total electric field at P will be along the y-axis and will have a

mag-nitude given by:

E=kQ

R2

sinφ/2

There are three special cases to Eq.20.43:

(1) φ = 0 (Point charge)

When we apply the limiting case lim

φ →0 [sin(φ/2)/(φ/2)] = 1,we get:

E= kQ

(2) φ = π (Half a circle of radius R)

When we substitute with sin(π/2)/(π/2) = 2/π, we get:

E= 2kQ

(3) φ = 2π (A ring of radius R)

When we substitute with sinπ = 0, we get:

This is an expected result, since we shall see that Eq.20.50gives E = 0 when P

is at the center of the ring, i.e when a = 0.

20.4.3 The Electric Field of a Uniformly Charged Ring

Assume that a ring of radius R has a uniformly distributed total positive charge Q,

see Fig.20.18 Also, assume there is a point P that lies at a distance a from the center

of the ring along its central perpendicular axis, as shown in the same figure

Fig 20.18 A ring of radius R

having a uniformly distributed

positive charge Q The figure

shows the electric field d→Eat

an axial point P due to a

segment of charge dq The

horizontal components will

cancel each other, and the total

field will be along the z-axis

P

d E

z

d q

θ

r

R

d E⊥

dE z

To find the electric field at P, we first calculate the magnitude of the electric field

dE at P due to this segment of charge dq as follows:

dE = k dq

This field has a vertical component dEz = dE sin θ along the z-axis and a component

dE⊥perpendicular to it, as shown in Fig.20.18 The perpendicular component created

at P by any charge segment is canceled by a symmetric charge segment on the opposite

side of the ring Thus, the perpendicular components of all of the charge segments

sum to zero Using r =√R2+ a2and sinθ = a/r, the vertical component will take

the form:

dEz = dE sin θ = k dq

r2

a

r = ka dq

(R2+ a2)3/2 (20.48) The total electric field at P due to all segments of the ring is given by the integration

of the z-component as follows:

E= dEz= ka dq

(R2+ a2)3/2

(R2+ a2)3/2



dq

(20.49)

Since

dq represents the total charge Q over the entire ring, then the total electric field at P will be given by:

E= (R2+ a kQa2)3/2 (20.50)

This formula shows that the field is zero at the center of the ring, i.e., at a = 0.When point P is very far from the ring, i.e., a  R, then we can neglect R2in the denominator

of Eq.20.50and get E ≈ kQ/a2 This form resembles the one we got for a point

charge

20.4.4 The Electric Field of a Uniformly Charged Disk

Assume that a disk of radius R has a uniform positive surface-charge density σ Also, assume that a point P lies at a distance a from the disk along its central perpendicular

axis, see Fig.20.19

To find the electric field at P, we divide the disk into concentric rings, then calculate the electric field at P for each ring by using Eq.20.50, and finally we can sum up the contributions of all the rings

Fig 20.19 A disk of radius R

has a uniform positive surface

charge densityσ The ring

shown has a radius r and radial

width dr The total electric

field at an axial point P is

directed along this axis

r

d r

R

P

a

z

Charge per unit area σ

E

Ring

Disk

Figure20.19shows one such ring, with radius r, radial width dr, and surface area

dA = 2πr dr Since σ is the charge per unit area, then the charge dq on this ring is:

Using this relation in Eq.20.50, and replacing E with dE, R with r, and Q with

dq = 2π rσ dr, then we can calculate the field resulting from this ring as follows:

dE=(r2+ a ka2)3/2 (2π rσ dr) = πkσa (r22r dr + a2)3/2 (20.52)

To find the total electric field, we integrate this expression with respect to the

variable r from r = 0 to r = R This gives:

E= dE = πkσ a

R



0

(r2+ a2) −3/2 (2r dr) (20.53)

To solve this integral, we transform it to the form

u n du = u n+1/(n + 1) by setting

u = r2+ a2, and du = 2r dr Thus, Eq.20.53becomes:

E = πkσ a

R



0

(r2+ a2) −3/2 (2r)dr = πkσ a

u =R2+a2

u =a2

u −3/2 du

= πkσ a

u −1/2 −1/2u =R

u = a2 = πkσa

(R2+ a2) −1/2

a−1

−1/2

 (20.54)

Rearranging the terms, we find:

E = 2πkσ



1−√ a

R2+ a2



(20.55)

Using k = 1/4π◦ , where ◦is the permittivity of free space, it is sometimes prefer-able to write this relation as:

E= σ

2◦



1−√ a

R2+ a2



(20.56)

We can calculate the field when point P is very close to the disk (the near-field approximation) by assuming that R  a, or by assuming the disk to be an infinite sheet when R → ∞ while keeping a finite In both cases, the second term between

the two brackets of Eq.20.56approaches zero, and the equation is reduced to:

E= σ

2◦

⎧

⎪

⎪

(Points very close to the disk)

or (Infinite sheet)

⎫

⎪

The concept of electric field lines was introduced by Faraday as an approach to help

us visualize electric fields

Spotlight

An electric field line is an imaginary line drawn in such a way that the direction

of its tangent at any point is the same as the direction of the electric field vector

→

Eat that point, see Fig.20.20

Since the direction of an electric field generally varies from one point to another, the electric field lines are usually drawn as curves, see Fig.20.20

Fig 20.20 The direction of

the electric field at any point is

the tangent to the electric field

line at this point

P

Q

Electric field

at point P

Electric field at

point Q

Electric field line

The relation between electric field lines and electric field vectors is as follows:

Spotlight

• The electric field vector →

E is tangent to the electric field line at any point

• The direction of the electric field line at any point is the same as the direction of the electric field

• The number of electric field lines per unit area, measured in the plane of the lines, is proportional to the magnitude of →

E Thus, the electric field

lines are closer together when the electric field is strong, and far apart when the field is weak

The rules for drawing electric field lines are as follows:

Spotlight

• Electric field lines must emerge from a positive charge and end on a neg-ative charge For a system that has an excess of one type of charge, some lines will emerge or end infinitely far away

• The number of lines emerging from a positive charge or ending at a negative charge is proportional to the magnitude of the charge

• Electric field lines cannot cross each other

The above rules are used in the six cases shown in Fig.20.21