Study Guide
Example Calculate the Electric Field
-
University:
Kentucky State University -
Course:
PHY 212 | General Physics II Academic year:
2019
-
Views:
405
Pages:
37
Author:
Arielle Armstrong
Example 4 : calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
R
y
x Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
d
y
R
Start with our usual equation:
dq
dE = k 2
r
Pick an infinitesimal dq of charge.
x
dq subtends an arc length ds, and an angle d.
What is the charge dq?
dq = charge per length of arc length of the arc
dq = ds Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
d
y
xdE′
dq′
R
dE
Draw the dE due to the dq, and
show its components.
Do you see any helpful symmetry?
Pick a dq′ horizontally across the arc from dq. The xcomponents of dq and dq′ will cancel. Because of this
symmetry, Ex = 0
Each dEy points downward so Ey will be negative. Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
dq
y
x
R
dE
Recall that dq and ds are
infinitesimal. dq is located at an
angle along the semicircle from
the negative y-axis.
is also one of the angles in the
vector triangle.
dEy = -dE sin Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
dq
y
x
R
dE
An arc of a circle has a length
equal to the circle radius times
the angle subtended (in radians):
ds = R d
Also,
=
charge on arc Q Q
length of arc 1 2R R
2 +Q
dq
ds
dq
y
x
R
dE
Let’s summarize what we have
done so far.
dq
dE = k 2
r
dq = ds
Q
=
R
ds = R d
Every dq is a distance R away from the arc center:
Q
R d
ds
Q d
R
dE = k 2 = k
=k
2
R
R
R 2
dEy = -dE sin
r=R +Q
dq
ds
dq
y
x
kQ
Ey = - 2
R
kQ d
dE y = sin
2
R
R
dE
kQ
arc sin d = - R 2
kQ d
E y = dE y = sin
2
R
arc
arc
kQ
0 sin d = + R 2 cos 0
kQ
kQ
2kQ
Ey = +
cos cos 0 = +
1 1 = 2
2
R
R
R 2 Review of Lecture 1, Part I
Coulomb's Law:
1 q1q 2
F =
,
2
12 4πε 0 r12
r12
+
-
q1
q2
Coulomb’s Law quantifies the force between charged particles. Review of Lecture 1, Part I
Electric field:
• charges create electric fields
• electric field of a point charge:
q
magnitude: E=k 2
r
direction: away from + ,
(Faucet)
towards
-
(Sink)
• force felt by charge q in electric field: F= qE
positive charge feels force in direction of E
negative charge feels force in direction opposite to E Today’s lecture:
Electric field due to a continuous distribution of charge.
We need to calculate electric field of a charge distributed along some line, over some surface, or
throughout some volume.
Problem:
• our equations for the Coulomb force and the electric
field hold for point charges only
Solution:
• decompose extended object into small charge
elements (chunks)
• calculate electric field for each element
• sum up (integrate) contributions of all elements to
obtain total electric field Volume Charge Density
• When a charge Q is distributed evenly throughout a
volume V, the Volume Charge Density is defined as:
ρ ≡ (Q/V) (Units are C/m3)
Surface Charge Density
• When a charge Q is distributed evenly over a surface
area A, the Surface Charge Density is defined as:
σ ≡ Q/A (Units are C/m2)
Linear Charge Density
• When a charge Q is distributed along a line ℓ , the
Linear Charge Density is defined as:
λ ≡ (Q/ℓ) (Units are C/m) Charge Density
l = Q/L
• Line of charge:
charge per unit length = l
• Sheet of charge:
charge per unit area = s
• Volume of charge:
charge per unit volume = r
s = Q/A
r = Q/V
When a charge Q is distributed
uniformly (evenly), then the
charge density (l , s , r ) is
constant Example 1: A rod of length L has a uniformly distributed total
negative charge -Q. Calculate the electric field at a point P
located a distance d below the rod, along an axis through the
center of and perpendicular to the rod.
-Q
L
Starting equation: dE = k
dq
r2
The equation says:
d
P
(1) pick a dq of charge somewhere in the distribution
(2) draw in your diagram the dE due to that dq
(3) draw the components of dE
(4) for each component, check for simplifications due to symmetry,
then integrate over the charge distribution. Calculate the electric field at a point P.
y
-Q
L
dq
Starting equation: dE = k 2
r
x
dq dE
d
P
Pick a dq (best to not put
it at either end or in the
middle).
Draw the dE due to the dq.
Now draw the components.
Do you see why symmetry tells me that Ex = 0? Calculate the electric field at a point P.
y
-Q
L
x
Ex = 0, so calculate Ey
x
dq dE
d
P
First, label an angle in
the vector diagram.
dEy = +dE sin
To find sin , we need the x-coordinate of dq. If dq is at an
arbitrary position along the x-axis, what is a good name for its
coordinate? That’s right, we’ll call it x.
The diagram is getting rather “busy,” but we are almost done
with it; don’t panic!. Calculate the electric field at a point P.
y
-Q
L
x
x
dq dE
d
P
To find sin , look at the green triangle. The sides have length
x and d, and hypotenuse r, where
r = x2 d2
From the green triangle, we see that sin = d / r. Calculate the electric field at a point P.
y
-Q
L
x
x
dq dE
d
P
Now we start to put things together:
dq
dq d
d dq
dq
dE y = dE sin = k 2 sin = k 2 = k 3 = kd
2
2 3/2
r
r r
r
x d )
To find Ey we simply integrate from one end of the rod to the
other (from –L/2 to L/2). Calculate the electric field at a point P.
y
-Q
L
x
x
dq dE
d
P
Ey =
L2
L 2
k
d dq
x
2
d
)
2 3/2
But wait! We are integrating over the rod, which lies along the
x-axis. Doesn’t there need to be a dx somewhere? Calculate the electric field at a point P.
y
-Q
x
dq
x
I removed un-needed “stuff” from the figure.
dq is a tiny bit (chunk) of charge on the uniformly charged rod.
If the charge is uniformly distributed, then the amount of
charge per length of rod is
(charge)
Q
(linear charge density) =
or
l=
(length)
L
Thus, dq = l dx, where dx is an infinitesimal length of rod that
lies along the x-axis Calculate the electric field at a point P.
y
dx
-Q
x
x
dq
We can take l outside the integral
because the charge is uniformly
distributed, so l must be constant.
Ey =
L2
L 2
dq
kd
Q
E y = kd
L
x
2
d
dx
L2
L 2
x
)
2 3/2
2
d
=
L2
L 2
)
2 3/2
kd
l dx
x
2
d
)
2 3/2
= kd l
dx
L2
L 2
x
2
d
)
2 3/2
The physics of the problem is all
done. The rest is “just” math. Example 2 : A rod of length L has a uniform charge per unit
length l and a total positive charge Q. Calculate the electric field
at a point P along the axis of the rod a distance d from one end.*
y
P
x
d
L
For the given x- and y-axes, let’s put the origin at P.
As in the previous example, we realize we will need to calculate
the linear charge density on the rod.
Q
l=
and Q = lL
L
*Assume the rod has negligible thickness. Example: A rod of length L has a uniform charge per unit length
l and a total positive charge Q. Calculate the electric field at a
point P along the axis of the rod a distance d from one end.
y
dE
dq
P
d
x
L
dq
We start with our usual equation: dE = k 2
r
The equation says pick a dq of charge, so do it!
Because dq is positive, its contribution to the electric field
points away from the rod. y
dE
x
dq
P
d
x
L
The electric field at point P has no y-component (why?).
Therefore, Ey=0.
The rod has
no thickness.
The infinitesimal charge dq is a distance x away from the origin.
dq
l dx
l dx
= -k 2
Therefore dE x = -k 2 = -k
2
x
x
x
The – sign is
important! y
dE
x
dq
P
d
x
L
Now simply integrate over the rod.
E x = dE x =
rod
dL
d
d L dx
l dx
-k 2 = -kl
d
x
x2
d d L)
1
1
1
E x = -kl = kl
= kl
xd
dL d
d d L)
dL
klL
Ex = d d L) y
dE
x
dq
P
d
x
L
The problem asks for the electric field at point P, thus the
complete answer is:
E y =0
klL
Ex = d d L)
or
klL ˆ
E=i
d d L)
How to validate your calculation? Use your intuition!
1. If d >> L, then E = ?
2. If d goes to 0, then E goes to ? Example 3 : A ring of radius a has a uniform charge per unit
length and a total positive charge Q. Calculate the electric field at
a point P along the axis of the ring at a distance x0 from its center.
P
x0
x Example: A ring of radius a has a uniform charge per unit
length and a total positive charge Q. Calculate the electric field
at a point P along the axis of the ring at a distance x0 from its
center.
y
Let’s add a y-axis to the
figure.
dq
r
a
P
x0
x
dE
Starting equation:
dq
dE=k 2
r
Pick a dq of charge. Let’s put it on the y-axis for now.
Show the dE due to that positive dq.
We’ll need r and later. y
dq
r
a
dE′
x0
r
dq′
P
dE
x
Show the x- and ycomponents of dE. There
may also be a z-component,
which we’ll leave out
because it is difficult to draw
and visualize.
Consider the dq’ on the ring where it is intersected by the
negative y-axis.
dq’ gives rise to dE’ at P. Show the components of dq’.
All points on the ring are the same distance r from point P.
Also, x0 and are the same for all points on the ring.
The y-components cancel pairwise! Same for the
z-components (not shown). Ey = Ez = 0. y
Back to our starting equation:
dq
r
a
P
x
x0
dE x = +dE cos
dE
From the diagram:
r=
dq
dE=k 2
r
x a
2
0
2
x0
cos =
r
For a given x0, r is a constant
for all points on the ring.
Only works because all Ex are in same direction.
dq x 0
x0
E x = dE x = k 2 = k 3
r r
r
ring
ring
kx 0 Q
x0
ring d q = k r 3 Q = x 2 a 2 3/2
0 ) y
Thus the complete answer:
a
P
x0
or
E=
x
E
kx 0 Q
2
2
x
a
0 )
ˆi
3/2
Ex =
x
kx 0 Q
2
0
a
)
2 3/2
E y = Ez = 0
These equations are only
valid for P along the x-axis!
How to validate your calculation?
1. If x0 >> a, then E = ?
2. If x0 = 0, then E = ? Symmetry..! Example 4 : calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
R
y
x Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
d
y
R
Start with our usual OSE.
dq
dE = k 2
r
Pick an infinitesimal dq of charge.
x
dq subtends an arc length ds, and an angle d.
What is the charge dq?
dq = charge per length of arc ) length of the arc )
dq = lds Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
d
y
xdE′
dq′
R
dE
Draw the dE due to the dq, and
show its components.
Do you see any helpful symmetry?
Pick a dq′ horizontally across the arc from dq. The xcomponents of dq and dq′ will cancel. Because of this
symmetry, Ex = 0
Each dEy points downward so Ey will be negative. Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
dq
y
x
R
dE
Recall that dq and ds are
infinitesimal. dq is located at an
angle along the semicircle from
the negative y-axis.
is also one of the angles in the
vector triangle.
dEy = -dE sin Example: calculate the electric field at “center” of semicircular
line of uniformly-distributed positive charge, oriented as shown.
+Q
dq
ds
dq
y
x
R
dE
An arc of a circle has a length
equal to the circle radius times
the angle subtended (in radians):
ds = R d
Also,
l=
charge on arc ) = Q = Q
length of arc ) 1 2R ) R
2 +Q
dq
ds
dq
y
x
R
dE
Let’s summarize what we have
done so far.
dq
dE = k 2
r
dq = l ds
Q
l=
R
ds = R d
Every dq is a distance R away from the arc center:
Q
R d )
l ds
Q d
R
dE = k 2 = k
=k
2
R
R
R 2
dEy = -dE sin
r=R +Q
dq
ds
dq
y
x
kQ
Ey = - 2
R
kQ d
dE y = sin
2
R
R
dE
kQ
arc sin d = - R 2
kQ d
E y = dE y = sin
2
R
arc
arc
kQ
0 sin d = + R 2 cos 0
kQ
kQ
2kQ
Ey = +
cos cos 0 = +
1) 1) = 2
2
R
R
R 2
Example Calculate the Electric Field
of 37
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