Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 10
Find E at A where OA = a.
1. E A
= k q 1
a 2 correct
2. E A
= k q 1
2 a 2
3. E A
= k q 1
b 2
4. E A
= k q 1
c 2
5. E A
= 0
6. E A
= k q 1
3 a 2
q 1
7. E A
= k √
2 a 2
8. E A
= k 2 q 1
a 2
9. E A
= k 3 q 1
a 2
10. E A
= k 4 q 1
a 2
Explanation:
Pick a Gaussian surface (sphere since we
are in spherical symmetry) center at the point
charge and of radius a. This surface contains
only the point charge, so q encl = q 1 . The
formula for E gives
8. E B
= k q 1 − q 2
b 2
9. E B
= k 3 q 1
b 2
10. E B
= k 4 q 1
b 2
Explanation:
For an electrostatic situation, inside of a
conductor, there is no charge; i.e., q inside = 0.
Also, ⃗ E inside = 0 and there is no flux inside,
Φ inside = 0.
Thus
E B
= 0 .
Notice also that since the electric field at B
is zero, the total enclosed charge is zero, or
q 1 + q ′ 2 = 0. Therefore
q ′ 2 = −q 1 .
This verifies that the charge on the inner
surface of a conducting shell is −q 1 , where
q 1 is the charge is the charge enclosed by the
shell.
022 (part 3 of 3) 10 points
Find E at C, where OC = c.
1. E C
= 0
E A
= k q 1
a 2 .
021 (part 2 of 3) 10 points
Find E at B, where OB = b.
1. E B
= 0 correct
2. E B
= k q 1
a 2
3. E B
= k q 1
b 2
4. E B
= k q 1
2 b 2
5. E B
= k q 1
c 2
6. E B
= k q 2
2 b 2
7. E B
= k q 1 + q 2
√
2 b 2
2. E C
= k q 1
a 2
3. E C
= k q 1 + q 2
b 2
4. E C
= k q 1 − q 2
2 a 2
5. E C
= k q 1
c 2
6. E C
= k q 1
2 b 2
7. E C
= k q 1 + q 2
c 2
8. E C
= k q 1 − q 2
c 2
9. E C
= k 3 q 1
c 2
10. E C
= k 4 q 1
c 2
Explanation:
correct