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Unformatted text preview: s case, the negative solution refers to a point to the right of the negative charge. This
puts our field point between the charges, which is a scenario we’ve already dismissed.
Therefore, our field point P must be located 1.8 m to the left of the negative charge. Solutions will be posted at www.prep101.com/solutions 8 ©Prep101 Phys1026 April Exam Solutions Question 2:
Solution:
For A:
r
r
r
k (Q1  Q 2 )
8.99 × 10 9 Nm 2 C 2 × 4.0 × 10 −7 C
ˆ
ˆ
E p = E p1 + E p2 =
x=
x
( d )2
( 0.15 m )2
2
2
r
5
1
ˆ
E p = 6.39 × 10 x N ⋅ C (or V ⋅ m 1 ) ( )( ) For B:
r
r
ˆ
Fp = qE p = 6.4 × 10 5 N ⋅ C −1 x × 1.0 × 10 −10 C
r
ˆ
Fp = 6.4 × 10 5 N x ( )( ) Question 3:
Solution: Answer (e)
Since B is repelled by A, we know that ball B is positively charged. However, C may be
attracted to B either because it is negatively charged, or because of an induced charge
making its left side slightly more negative and its right side slightly more positive.
Question 4:
Solution: After simplification gives: Solutions will be posted at www.prep101.com/solutions 9 ©Prep101 Phys1026 April Exam Solutions Question 5:
Solution:
Consider a differential element of length d l = Rd θ , which makes an angle θ with the xaxis, as shown in the figure. The amount of charge it carries is dq = λ d l = λ Rd θ The contribution to the electric field at O is: Integrating over the angle from − θ o to + θ o , we have We see that the electric field only has the xcomponent, as required by a symmetry
argument. If we take the limit θ o → π , the arc becomes a circular ring. Since sin π = 0 ,
the equation above implies that the electric field at the center of a nonconducting ring is
zero. This is to be expected from symmetry arguments. On the other hand, for very small
θ o , sin θ o ≈ θ o and we recover the pointcharge limit: where the total charge on the arc is
Q = λ l = λ (2 R θ o ) Answer 3.6 The net electric field at point D has components from the charges at A, B,
C:
k (−Q) kQ
EB =
= 2 N/C [DOWN]
d2
d
k (−Q) kQ
EC =
= 2 N/C [LEFT]
d2
d
k (Q)
kQ
N/C
=
EA =
2
2
2
2d 2
d +d ( ) Solutions will be posted at www.prep101.com/solutions 10 ©Prep101 Phys1026 April Exam Solutions kQ
kQ
cos 45° N/C; E Ay =
sin 45° N/C
2
2d
2d 2
The net electric field is the sum of the components:
kQ
kQ kQ ⎛ cos 45° ⎞
kQ
E at D , x =
cos 45° − 2 = 2 ⎜
− 1⎟ = −0.65 2 N/C
2
d
d⎝ 2
d
2d
⎠
kQ
kQ kQ ⎛ sin 45° ⎞
kQ
E at D , y =
sin 45° − 2 = 2 ⎜
− 1⎟ = −0.65 2 N/C
2
d
d⎝ 2
d
2d
⎠
kQ
kQ ⎞
⎛
E at D = ⎜ − 0.65 2 ,−0.65 2 ⎟ N/C
d
d⎠
⎝
kQ
kQ ⎞
⎛
Field at D = E at D = ⎜ − 0.65 2 ,−0.65 2 ⎟ N/C
d
d⎠
⎝
E Ax = GAUSS’ LAW
PRACTICE QUESTIONS
Question 1:
Solution:
Recall that the flux is given as
Φ E = E ⋅ A = EA cos θ where θ is the angle between the vector area A and the electric field E. In order for the
flux to be at a maximum, we must have cos θ = 1 , which will occur when the vector area
of the loop is aligned parallel to the field.
In the case of...
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 Winter '14
 Prof.PaulineBarmby
 Physics

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