Question

A straight rod of length =22.80 cm carries a uniform charge density =1.80×10^−6 C/m. The rod is located along the ‑axis from 1=0.00 to 2=. The Coulomb force constant is =8.99×10^9 N·m2/C2.

A vertical line of positive charge of length capital L. A labeled point P is located above the line of charge along a line coaxial with the line of charge. The distance from the base of the line of charge to point P is y subscript zero, with y sub zero greater than L.

Find the expression for the electric field along the y‑axis Ey at a point P. What is the magnitude of the electric field at y0=60.00 cm?

Help is greatly appreciated!!

Answer 1

The value of the length of the rod is
L
=
22.80
c
m
The value of the linear charge density on the rod is
λ
=
2.85
×
10
−
6
m
−
1
From the higher end of the rod, the distance of the test point is
y
∘
=
65
c
m
According to Coulomb's Law,

The value of the electric field due to the small segment of the rod is,

d
E
=
k
d
q
r
2
d
r
d
E
=
k
λ
L
r
2
d
r
dE =
k
d
q
r
2
dr dE =
k
λ
L
r
2
dr

Where k is the electrostatic constant.

The distance of the lower end from the origin is
y
1
=
0
y
1
=
0

The distance o the higher end of the rod from the origin is

y
2
=
L
y
2
=
L

Thus, the distance of the test point from the lower end is,

y
′
=
y
2
+
y
∘
y
′
=
y
2
+
y
∘

The net electric field at the test charge due to the linear charge density of the rod is,

E
=
∫
y
′
y
∘
k
λ
L
r
2
d
r
E
=
k
λ
L
∫
y
′
y
∘
1
r
2
d
r
E
=
k
λ
L
[
−
1
r
]
y
′
y
∘
E
=
k
λ
L
[
−
1
y
′
+
1
y
∘
]
E =
∫
y
∘
y
′
k
λ
L
r
2
d
r
E =kλL
∫
y
∘
y
′
1
r
2
d
r
E =kλL
[
−
1
r
]
y
∘
y
′
E =kλL
[
−
1
y
′
+
1
y
∘
]


Substituting the known values,

E
=
8.99
×
10
9
×
2.85
×
10
−
6
×
22.8
×
10
−
2
[
−
1
(
22.80
+
65
)
×
10
−
2
+
1
65
×
10
−
2
]
E
=
2.33
×
10
3
N
−
1
E =8.99×
10
9
×2.85×
10
−
6
×22.8×
10
−
2
[
−
1
(
22.80
+
65
)
×
10
−
2
+
1
65
×
10
−
2
]
E =2.33×
10
3
N
−
1


Thus, the electric field at the test point is
2.33
×
10
3
N
−
1
2.33
×
10
3
N
−
1