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

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 26
4. Cannot be determined
V r = V R + ∆V
= k Q ∫ r
R − Explanation:
E dr
We know that the potential due to a collection
of N point charges is given by
R
= k Q R − k Q ∫ r
R 3 r dr , from Eq. 6
R
= k 2 Q
2 R + k Q (
R 2
2 R 3 − r 2)
V = 1 ∑
N q i
4 π ɛ 0 r
i=1 i
= k 3 Q
2 R − k Q
2 R 3 r2 = 1 ( q
4 π ɛ 0 a + −q )
= 0
a
8. Z
= k Q (
3 − r
2 ) ,
2 R
where r < R .
9. M
Explanation:
The electric potential for R < r with
the sphere conducting and/or uniformly
V
V ∝ 1 r
non-conducting: In the region outside the
uniformly charged non-conducting sphere, we
G.
have the same conditions as for the conducting
sphere when applying the definition for
the electric potential; therefore,
0 R
r
∫ r
V = − E dr
keywords:
∞
∫ r
dr
Finding Zero Potential
= −k Q
∞ r 2
25:06, trigonometry, multiple choice, < 1 min,
= k Q fixed.
, where R < r . (4)
051 (part 1 of 4) 10 points
r
All of the charges shown are of equal magnitude.
The electric potential for r < R with
the sphere uniformly non-conducting:
Because the potential must be continuous at
r = R , we can use this expression to obtain
the potential at the surface of the sphere; i.e.,
−q +q
the potential at a point on the conducting
sphere is V = k Q a a
r
From Part 2 we found that the electric field
inside an uniformly charged non-conducting
(a)
sphere is
What is the electric potential E at the origin
Assume zero potential at infinity.
E = k Q r , where r < R . (6)
R3 We can use this result in the definition for
the electric potential to evaluate the potential
difference ∆V = V r − V R (where V R = k Q R as
shown in Eq. 4) at some interior point of the
1. zero correct
2. positive
3. negative
sphere, so