CH 23

1
CH 23 : CONTINUOUS CHARGE DISTRIBUTIONS AND GAUSS’S LAW
KHALIL BAZZ 55399534
Multiple Choice Questions
1. A uniform line charge density λ = 10.0 nC/m extends from x = 0 to x = 10 m.
The magnitude of the electric field at x = 12 m is:
a) 3.86 N/C b) 9.62 N/C c) 15.3 N/C d) 18.8 N/C e) 37.5 N/C
E=kdq
r
E=kλ dx
x
E=kλx dx
E=−kλ 1
x
E=−(9×10
)(10×10 )1
12−1 2=37.5 N/C
2. A linear charge of uniform density -20 nC/m is distributed along the y axis from
y = 2.0 cm to y = 5.0 cm. The magnitude of the electric field at the origin (in N/C) is:
a) 1800 (ȷ̂) b) 5400 (ȷ̂) c) 5400 (−ȷ̂)
d) 6300 (ȷ̂) e) 1800 (−ȷ̂)
E=kdq
r
E=kλ dy
y
E=kλy dy
E=−kλ 1
y
.
.
E=−(9×10
)(20×10 )1
0.05−1 0.02=5400 N/C

2
CH 23 : CONTINUOUS CHARGE DISTRIBUTIONS AND GAUSS’S LAW
KHALIL BAZZ 55399534
3. For the charged rod shown in the figure below, if Q = 1 nC, L = 70 cm, a = 30 cm.
The electric field (in N/C) at point P is:
a) 30 b) -30 c) 9
d) -9 e) 60
λ=Q
L=1×10 0.7 =1.43×10
C
E=kdq
r
E=kλ dx
x
E=kλx dx
E=−kλ 1
x
.
.
E=−(9×10
)(1.43×10 )1
1−1 0.3=30 N/C
4. A thin rod of length L carries a total charge Q
distributed uniformly along its length. Determine
the electric field at point P. (a > 0).
a)
()
b)
()
c)
()
d)
()
e)
()
E=kdq
r
E=kλ dx
x
E=kλx dx
E=−kλ 1
x
=−k(Q
L)1 a+L−1 a
E=−kQ
L −L a(a+L)=kQ a(a+L)
L
a
P

3
CH 23 : CONTINUOUS CHARGE DISTRIBUTIONS AND GAUSS’S LAW
KHALIL BAZZ 55399534
5. Two long straight parallel lines of charges Q
1
and Q
2
, carry positive charge per unit
lengths of λ
1
and λ
2
, respectively. λ
2
> λ
1
. The locus of points where the electric field
is zero in this case is:
a) At a point halfway between the lines.
b) Along a line between the lines closer to line #1 than line #2.
c) Along a line between the lines closer to line #2 than line #1.
d) Along line #1.
e) Cannot be determined
6. A uniformly charged rod, 2.0 m long, carries a charge with a linear charge density
equal to 3.0 nC/m is bent to form a semicircle. The magnitude of the electric field
(in N/C) at the center of the circle is:
a) 64 b) 133 c) 48 d) 84 e) zero
L=Rθ → R=L
θ=2 π=0.64 m
E=kdq
r
sinθ
E=kλ R dθ
R
sinθ
E=k λ
Rsinθ dθ
E=k λ
R [−cosθ]
E=−(9×10
) (3 × 10 )
0.64 [cosπ−cos0]=|−84.4|=84.4 N/C
7. A charge of uniform density of 3.5 nC/m is distributed along a circular arc as shown.
The magnitude of the electric field in (N/C) at point P is:
a) 76.5 b) zero c) 126 d) 31.5 e) 63
E=kdq
r
cosθ
E=kλ R dθ
R
cosθ
E=k λ
Rcosθ dθ
E
x
E
x
E
y
E
y
0
π

4
CH 23 : CONTINUOUS CHARGE DISTRIBUTIONS AND GAUSS’S LAW
KHALIL BAZZ 55399534
E=k λ
R [sinθ]
E=(9×10
) (3.5 × 10 )
0.5 sin (7 6π) − sin (5 6π)
E=|−63|=63 N/C
8. A uniformly charged thin rod, linear charge density λ (Cm
-1
) is bent into a semicircle
of radius R, as shown. The magnitude of the electric field at the center P of the
semicircle is:
a) 2λπϵ
°
 b) 4λπϵ
°
 c) λπϵ
°
 d) λ2πϵ
°
R
 e) λ2πϵ
°
E=kdq
r
sinθ
E=kλ R dθ
R
sinθ
E=k λ
Rsinθ dθ
E=k λ
R [−cosθ]
E=−1
4πϵ
°
λ
R[cosπ−cos0]=2λ 4πϵ
°
R=λ 2πϵ
°
R
9. A uniformly charged insulang rod of length 14.0 cm is bent into the shape of a
semicircle as shown in the figure. The rod has a total charge of -7.50 µC. what is the
magnitude and direction of the electric field at O, the center of the semicircle?
a) 2.1 x 10
7
N/C (positive x-axis)
b) 2.1 x 10
7
N/C (negative x-axis)
c) 2.1 x 10
7
N/C (positive y-axis)
d) 2.1 x 10
7
N/C (negative y-axis)
e) zero
L=Rθ → R=L
θ=0.14 π=0.045 m
λ=Q
L=7.5×10 0.14 =5.36×10
C/m
+
+
+
+
+
E
x
E
x
E
y
E
y
0
π
x
y

5
CH 23 : CONTINUOUS CHARGE DISTRIBUTIONS AND GAUSS’S LAW
KHALIL BAZZ 55399534
E=kdq
r
cosθ
E=kλ R dθ
R
cosθ
E=k λ
Rcosθ dθ
E=k λ
R [sinθ]
E=(9×10
) (5.36 × 10 )
0.045 sin (3π 2) − sin (π 2)
E=|−2.1×10
|=2.1×10
N/C
10. A thin circular disk of radius R=30 cm is oriented in the yz-plane with its center as
the origin. The disk carries a total charge Q=+3 µC distributed uniformly over its
surface. Calculate the magnitude of the electric field due to the disk at the point
x=15 cm along the x-axis.
a) 3.32 x 10
5
N/C
b) 2.49 x 10
5
N/C
c) 1.99 x 10
5
N/C
d) 4.89 x 10
5
N/C
e) 9.95 x 10
5
N/C
E=kdq
r
cosθ
E=kσ (2πr dr)
(r +x ) . x
(r +x )
E=kσ 2π xr dr
(r +x )
.
E=kσ 2π x −1
(r +x )
.
E=−(9×10 )(3×10
π×0.3
)(2π)(0.15)1
(0.3 +0.15 )
−1
(0 +0.15 )
E=3.32×10 N/C
E
y
E
y
E
x
E
x