Problem #1
A uniformly charged conducting sphere of 1.2 m diameter has a surface charge density of 8.1 μC/m2. (a) Find the net charge on the sphere. (b) what is the total electric flux leaving the surface of the sphere.Answer:
(a) Field due to a uniformly charged thin spherical shell:
Q = σ(4πR2) = (8.1 x 10─6C/m2)(4π)(0.6 m)2
= 3.66 x 10─5 C
Q = 36.6 μC
(b) the total electric flux leaving the surface of the sphere is given by
φ = Q/ϵ0
= 3.66 x 10─5 C/8.85 x 10-12
φ = 4.1 x 10─6 Nm2/C
Problem #2
A point charge q1 = 4.00 nC is located on the x-axis at x = 2.00 nm and a second point charge q2 = ─6.00 nC is on the y-axis at y = 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?
Answer
Apply Gauss’s law to the spherical surface. Qenclose is the algebraic sum of the charges enclosed by the sphere.
(a) No charge enclosed so ΦE = 0
(b) the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius 1.50 m is given by
ΦE = q2/ϵ0
= ─6.00 x 10─9 C/8.85 x 10-12 N.m2
ΦE = ─678 N.m2/C
(b) the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius 2.50 m is given by
ΦE = (q1 + q2)/ϵ0
= (4.00 x 10─9 C ─ 6.00 x 10─9 C)/8.85 x 10-12 N.m2
ΦE = ─226 N.m2/C
Negative flux corresponds to flux directed into the enclosed volume. The net flux depends only on the net charge enclosed by the surface and is not affected by any charges outside the enclosed volume.
Problem #3
An isolated conductor of arbitrary shape has a net charge of +10 x 10─6 C. inside the conducting is a cavity within which is a point charge q = +3.0 x 10─6 C. what is the charge (a) on the cavity wall and (b) on the outer surface on the conductor?
Answer:
Known:
Q = charge on conductor on outside surface = +10 x 10─6 C
S’ = Gaussian surface onclosing cavity. φ = 0; E = 0 everywhere on S inside conductor
So 0 = q + qwall
(a) the charge on the cavity wall is given by
0 = q + qwall
qwall = ─ q = ─3.0 x 10─6 C
qenclosed = q + Q
= 3 x 10─6 C + 10 x 10─6 C = 13 μC
Problem #4
The electric field just above the surface of the charged conducting drum of a photocopying machine has a magnitude E of 2.3 x 105 N/C. What is the surface charge density on the drum?
Answer:
Known:
magnitude electric field E = 2.3 x 105 N/C,
The surface charge density on the drum given by is
EA = Q/ϵ0
σ = Q/Adrum = Eϵ0
= (2.3 x 105 N/C)( 8.85 x 10-12 N.m2)
σ = 2.04 x 10─6 C/m2 = 2,04 μC
Problem #5
Flux and conducting shells. A charged particle is held at the center of two concentric conducting spherical shells. Figure (a) shows a cross section. Figure shows a cross section. Figure (b) gives the net flux φ through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere. The scale of the vertical axis is set by φS = 5.0 x 105 N.m2/C. What are (a) the charge of the central particle and the net charges of (b) shell Aand (c) shell B?
(a) Calculate the charge of central particle.
qcentral = ϕcentral × ϵ0
Here, ϕcentral is center flux and ϵ0 is permittivity of free space.
Known:
ϕcentral = −9.0 × 105 N.m2/C and ϵ0 = 8.85×10−12 N.m2,
then,
qcentral = (−9.0 × 105 N.m2/C) × (8.85 × 10−12 N.m2)
qcentral = −7.965 μC
(b) Calculate the total charge enclosed by shell A.
Known:
ϕA = +4.0 × 105 N.m2/C and ϵ0 = 8.85×10−12 N.m2,
qA + (−7.965 μC) = (+4 × 105 N.m2/C) × (8.85 × 10−12 N.m2) = 3.54 x 10−6 C
qA = +11.505 μC
(b) Calculate the total charge enclosed by shell B.
Known:
ϕB = −2.0 × 105 N.m2/C and ϵ0 = 8.85×10−12 N.m2,
qB + (3.54 μC) = (−2.0 × 105 N.m2/C) × (8.85 × 10−12 N.m2) = −1.77 x 10−6 C
qA = −5.31 μC
Problem #6
Space vehicles traveling through Earth’s radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metal satellite 1.3 m in diameter accumulates 2.4 μC of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.
Answer:
Known:
accumulates of charge in one orbital revolution, Q = 2.4 μC = 2.4 x 10−6 C
diameter spherical d = 1.3 m, or radius, r = 0.65 m
(a) the resulting surface charge density given by is
σ = Q/Aspherical = (2.4 x 10−6 C)/(4π x 0.652 m2) = 0.452 μC/m2
(b) the magnitude of the electric field just outside the surface of the satellite, due to the surface charge given by is
E = kQ/r2
= (9.0 x 109 Nm2.C−2)(2.4 x 10−6 C)/(0.65 m)2
E = 5.1 x 104 N/C
Note that if this charge is on the outside of the satellite it should form a faraday cage and the inside should not have any field acting.
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