Download Physics 122-02 Test 2 1. Two point charges are placed on the x axis

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Physics 122-02 Test 2
1. Two point charges are placed on the x axis, at x = a and x = −a. Find
an expression for the electric field (magnitude and direction) at all points on
the y axis (that is, all points of the form (x = 0, y, z = 0)) for the following
configurations:
(a) (4 pts.) A positive charge Q at x = a, and another positive charge Q at
x = −a.
Answer:
E=
Q
y
ŷ
2
2πǫ0 (a + y 2 )3/2
(b) (4 pts.) A negative charge −Q at x = a, and a positive charge Q at x = −a.
Answer:
E=
Q
a
x̂
2πǫ0 (a2 + y 2 )3/2
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2. A point charge q is located at the origin. Centered about the origin is a
conducting spherical shell of inner radius a and outer radius b.
(a) (1 pt.) What is the electric field E inside the conductor (a < r < b)?
Answer:
E=0
(b) (2 pts.) What is the total charge on the inner surface (r = a) of the shell, if
the shell is uncharged overall?
Answer:
−q
(c) (2 pts.) What is the total charge on the outer surface (r = b) of the shell, if
the shell is uncharged overall?
Answer:
+q
2
(d) (2 pts.) What is the total charge on the inner surface (r = a) of the shell, if
the shell carries an overall charge Q?
Answer:
−q
(e) (2 pts.) What is the total charge on the outer surface (r = b) of the shell, if
the shell carries an overall charge Q?
Answer:
Q+q
(f) (1 pt.) Suppose that the point charge q is moved slightly away from the
origin, but still remains within the cavity of the conducting shell r < a. Do
your answers to parts (a) through (e) change at all (yes or no)?
Answer:
no
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3. The electric potential V is given over all space by the expression
V (x, y, z) = 10.0(N/C)x − 4.0(N/C)y
(independent of z), where x and y are measured in meters (note that this means
V has units of Volts, as it should!).
(a) (3 pts.) What is the electric field E at each point in space?
Answer:
E(x, y, z) = −10.0(N/C)x̂ + 4.0(N/C)ŷ
(b) (4 pts.) A charge of magnitude +0.10C moves from the origin, (x, y, z) =
(0, 0, 0), to the point (x, y, z) = (1m, 0, 0). By what amount does its kinetic
energy change as a result? (Tell whether it increases, decreases, or stays the
same, and by how much if it does not stay the same.)
Answer:
Kinetic Energy decreases by 1.00 J
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Useful Formulae
2
F = qE
ǫ0 = 8.854 × 10−12 C 2 /(N m
e = 1.602 × 10−19 C
HR )
9
2
2
1/(4πǫ0 ) = 8.99 × 10 N m /C
ΦE =
E · n̂ dA = qenclosed/ǫ0
F = −∇U
E = −∇V
∇ = x̂
F=
1 q1 q2
r̂
4πǫ0 r2
U=
1 q1 q2
4πǫ0 r
1 q
r̂
4πǫ0 r2
V =
1 q
4πǫ0 r
E=
∂
∂
∂
∂
∂
1 ∂
∂
∂
1 ∂
1
+ ŷ
+ ẑ
= ρ̂
+ φ̂
+ ẑ
= r̂
+ θ̂
+ φ̂
∂x
∂y
∂z
∂ρ
ρ ∂φ
∂z
∂r
r ∂θ
r sin θ ∂φ
1J = 1N m
1N = 1kg m/s2
1V = 1J/C
cos = adj./hyp.
sin = opp./hyp.
tan = opp./adj.
Wx = W cos θ
Wy = W sin θ, for vector W making an angle θ measured
CCW. from the x axis
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