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=⎜2
⎜d
⎝ 2 ⎞ ⎛ kQ 2
⎟ +⎜ 2
⎟ ⎜d
⎠⎝ 2 ⎞
⎛ kQ 2
⎟ = 2⎜ 2
⎟
⎜d
⎠
⎝ 2 ⎞
kQ 2
⎟=2 2 N
⎟
d
⎠ Force = Fnet = 2 FBonA
kQ 2
N [x 45°y]
d2 (b) This is a conceptual question. If you wanted to, you could do some really tricky
math to obtain an equation for the coordinates of some general point in space where the
field is zero, but you won’t be able to solve it. The electric field is only zero in a few
Solutions will be posted at www.prep101.com/solutions 6 ©Prep101 Phys1026 April Exam Solutions special cases: at infinity; in the absence of all charges; or in certain very symmetrical
cases. Our particular case is none of these, so you should answer, NO. Something that
could tip you off to the answer is the fact that the question says “If so, how many points
are there, and where are they located? (Do not find the coordinates)”. Where are the
points, but don’t tell us where they are…that’s a contradiction.
Points where field is zero = NONE
Answer 2.5
kq1 q 2
∴ FE α q1, FE α q2, FE α 1/r2
2
r
a)
q1 → 2q1;
q2 → 2q2;
r=r
4kq1 q 2
→ 4(3.0 x 106N) = 1.2 x 105N
∴ FE →
r2
b)
uncharged sphere will ‘accept’ half of the charge from q1
kq q
∴ FE → ½ 12 2 → ½(3.0 x 106N) = 1.5 x 106N
r
c)
r → 3r
kq1 q 2
∴ FE →
→ (3.0 x 106N)/9 = 3.3 x 107N
(3r ) 2
Answer 2.6
For equilibrium, F13 = F23
kq1 q3
kq 2 2q3
=
 k, q3 cancel out
2
x
(2 − x) 2
1.6x105/x2 = 6.4x105/(2x)2
1/ x2 = 4/(2x)2
(2x)2 = 4x2
4 – 4x – x2 = 4x2
use quadratic formula to solve for roots
3x2 + 4x –4 = 0 FE = − b ± b 2 − 4ac
2a
− 4 ± 16 − 4(3)(−4)
x=
2(3)
x = 2.0 or x = 0.67
∴ the 3rd sphere must be 0.67m from the 1st sphere. x= Solutions will be posted at www.prep101.com/solutions 7 ©Prep101 Phys1026 April Exam Solutions ELECTRIC FIELDS
PRACTICE QUESTIONS
Question 1:
Solution:
In order for the net electric field to be zero, we need the fields due to the positive and
negative charges to be of equal magnitude, but directed in opposite directions. The field
due to the positive charge is going to point away from the positive charge; the field due to
the negative charge is going to point towards the negative charge. Based on this
knowledge alone, we know that our field point P cannot be between the charges. In this
region both the electric fields are going to be pointing to the left. Based on the magnitude
of the charges, we also know that the field point must be closer to the smaller charge,
since the field due to the larger charge is going to be larger in magnitude. As indicated in
the illustration, this means that the field point must be located to the left of the charges. Now, we can solve for the location of P by requiring that E + = E − : (6 .0 μ C )
1 (− 2 .5 μ C )
1
=
2
4πε o
4πε o (1m + d ) 2
d
Now we just need to solve for d:
2 .5 (1m + d ) 2 = 6 .0 d 2
1m + d = ± 6 .0
= ± 1 .55 d
2 .5 and, isolating d we find d= 1m
= 1.8m
+ 1.55 − 1 and d= 1m
= −0.39m
− 1.55 − 1 In thi...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
 Winter '14
 Prof.PaulineBarmby
 Physics

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