Unformatted text preview: Version 001 – Homework 3 – Hoﬀmann – (56605)
Calculate the potential diﬀerence between
point A and B.
Correct answer: 9.5 V. 4 5. VP = 0
2 ke Q 1 ke Q 2
+
a+b
c
2 ke Q 1 ke Q 1 ke ( Q 1 − Q 2 )
7. VP =
+
−
a+b
b
c
2 ke Q 1
8. VP =
a
2 ke Q 1 ke Q 2
9. VP =
−
a+b
c
ke Q 1 ke Q 2
10. VP =
−
a+b
b
Explanation:
Using the superposition principle, adding
the 3 concentric charge distributions; i.e., Q1
at a, −Q1 at b and Q1 + Q2 at c, gives
6. VP = Explanation:
Let : q = 2 C and
W = 19 J .
The work is
W = qV
W
19 J
V=
=
= 9. 5 V .
q
2C
Concentric Conductors 03 p2e1
010 10.0 points
Consider a solid conducting sphere with a
radius a and charge Q1 on it. There is a
conducting spherical shell concentric to the
sphere. The shell has an inner radius b (with
b > a) and outer radius c and a net charge
Q2 on the shell. Denote the charge on the
inner surface of the shell by Q2 and that on
the outer surface of the shell by Q2 .
Q1 , a b , Q2
P Q2 V= 2 ke Q 1 ke Q 1 ke ( Q 1 + Q 2 )
.
−
+
a+b
b
c Alternative Solution:
V =−
=−
=−
=− Er · dr , by symmetry, Er dr
c
∞
c
∞ Er dr − (a+b)/2 Er dr b ke ( Q 1 + Q 2 )
dr
r2
− Q2 , c Q1 Assume: The potential at r = ∞ is zero.
Find the potential VP at point P , which is
the midpoint between a and b.
1. VP =
2. VP =
3. VP =
4. VP =
rect 2 ke Q 1 2 ke Q 2
−
a+b
b
2 ke ( Q 1 − Q 2 )
a+b
2 ke Q 1
a+b
2 ke Q 1 ke Q 1 ke ( Q 1 + Q 2 )
−
+
cora+b
b
c = −ke (Q1 + Q2 ) 1
r (a+b)/2
b
c
∞ −ke Q1
= 1
r ke Q 1
dr
r2 (a+b)/2
b 2 ke Q 1 ke Q 1 ke ( Q 1 + Q 2 )
−
+
.
a+b
b
c keywords:
Charge and Concentric Spheres
011 10.0 points
A charge of 2 pC is uniformly distributed ...
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This note was uploaded on 09/25/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
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