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

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 22
The potential difference from O to A is
defined as
∆V = V A − V O = −
∫ A
O
⃗E · d⃗s .
We know that E ⃗ = (250 V/m) î . We need
to choose a path to integrate along. Because
the electric force is conservative, it doesn’t
matter which path we take; they all give the
same answer. There are two choices of path
for which the math is simple (see the figure
below.)
y
E
O
Path I:
II
I
(x, y)
B
A
I
x
which is the same as the result for the other
path.
keywords:
Potential Diagrams 02
25:04, calculus, multiple choice, > 1 min,
wording-variable.
047 (part 1 of 4) 10 points
Consider a sphere with radius R and charge
Q
Q
V A − V O = (V A − V B ) + (V B − V O ),
From O to B, ⃗ E and d⃗s are both along the
x-axis, so ⃗ E · d⃗s = E dx. From B to A, ⃗ E and
d⃗s are perpendicular, so ⃗ E · d⃗s = 0.
and the following graphs:
∝ 1 r
∫ B
V A − V O = −
= −
= −E
O
∫ x
0∫ x
O
⃗E · d⃗s −
E dx −
∫ A
B
∫ y
0
0 dy
dx = −E ∆x
= −(250 V/m) (0.2 m)
= −50 V .
⃗E · d⃗s
Q.
G.
0
R
∝ 1 r
r
The absolute value is
|∆V | = 50 V .
0
R
r
Path II: In this case, E ⃗ · d⃗s = E cos θ ds .
where cos θ = x ⇒ x = l cos θ .
l
∝ 1 r 2
∫ l
V A − V O = −E cos θ ds
O
= −E l cos θ
= −E x .
X .
0
R
r