Science > Physics > hellllllllllllllllllllllllllllp..PLZ life or death situation
| Topic: |
Science > Physics |
| User: |
"hype_chicky" |
| Date: |
10 Jan 2005 08:29:08 PM |
| Object: |
hellllllllllllllllllllllllllllp..PLZ life or death situation |
1. The electrostatic force between two point charges is 1.2 x 10^-2 N.
If the distance between them is doubled, the charge of one of the
point’s doubles and the charge of the other point triples. What will
be the force between them?
Kq1q2/r2
3. Two charges one charge +1.5 x 10^-2 C and the other charge -2.7 x
10^-5 C, are 20.0 cm apart. The positive charge is to the left of the
negative charge.
a) Draw a diagram showing the point charges and label a point Y that
is 5.0 cm away from the positive charge, on the line connect the
charge (field lines need not be drawn.)
b) b.) Calculate the electric field at point Y
4. Find the final speed of an electron, starting form rest, passing
between two parallel plates with a potential difference of 4.5 x 10^3
V. An electron mass of 9.1 x 10^-31 kg and a charge of 1.6 x 10^-19
C. Is this answer valid? Explain.
(yes it is valid because …)
5. Two parallel plates labeled W and X are separated by 5.2 cm. The
electric potential between the plates is 150V. An electron starts
from rest at time tW and reaches plate X at time tX. The electron
continues through the opening and reaches point P at time tp
(remember e = - 1.6 X 10^-19 C and the mass of an electron is 9.1 X
10^-31 kg)
a.) Sketch the speed-time graph on the axes below
b.) Determine the kinetic energy of the electron as it arrives at
plate X
SOLUTIONS THIS IS WHAT I HAVE THUS FAR:
For question 1:
K doesn't change, q1 is doubled so it's 2, q2 is tripled, so it's now
3 and r doubled, so the new force is 2*3*k/4 which is 3/2 of the old
one. So just multiply the 1.2 x 10^-2 N by 3/2. And the answer wud be
1.8x10^-2N
FOR QUESTION 3
For 3. I have no idea how to do the diagram I am like really lost L,
for the electric field at point Y I have E on point y = kq1/r^2 =
(9.0 x 10^9 Nm^2/c^2) (1.5 X10^-2)/ 0.05^2 and I got 5.4 X10^10N but
I feel like something is missing.
FOR QUESTION 4
Q deltaV /r = ˝ kmv^2
(1.6 X 10^ -19 C ) (4.5 X 10^3V) / 3,2 X10 ^-13 = ˝ x 9.1 x 10^-31 X
v^2
v^2 = 4.94 X 10^24
v = &.03 X 10^13
yes this is possible…but why?
FOR QUESTION 5
FE = Ek
Q delta V/r = 1/3 mv^2
1.6 x 10^-19 C (150V) / 0.05a = ˝ (9.1X10^-31 kg) (v^2)
V= 2.9 X 10^-23 m/s
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Posted at:
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| User: "Timo Nieminen" |
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| Title: Re: hellllllllllllllllllllllllllllp..PLZ life or death situation |
10 Jan 2005 09:07:02 PM |
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On Tue, 10 Jan 2005, hype_chicky wrote:
3. Two charges one charge +1.5 x 10^-2 C and the other charge -2.7 x
10^-5 C, are 20.0 cm apart. The positive charge is to the left of the
negative charge.=20
a)=09Draw a diagram showing the point charges and label a point Y that
is 5.0 cm away from the positive charge, on the line connect the
charge (field lines need not be drawn.)
b)=09b.) Calculate the electric field at point Y
[rearranged]
For 3. I have no idea how to do the diagram I am like really lost L,
for the electric field at point Y I have E on point y =3D kq1/r^2 =3D
(9.0 x 10^9 Nm^2/c^2) (1.5 X10^-2)/ 0.05^2 and I got 5.4 X10^10N but
I feel like something is missing.
That's (almost) fine for one charge. What about the effect of the other=20
charge?
For a diagram, surely it is just something like:
(+) 5 cm Y 15 cm (-)
Don't forget that the electric field is a vector quantity. Is the electric=
=20
field at Y due to the positive charge towards the left or right? The=20
electric field at Y due to the negative charge? So, do you add them or=20
subtract one from the other? In which direction is the total electric=20
field?
4. Find the final speed of an electron, starting form rest, passing
between two parallel plates with a potential difference of 4.5 x 10^3
V. An electron mass of 9.1 x 10^-31 kg and a charge of 1.6 x 10^-19
C. Is this answer valid? Explain.
=20
(yes it is valid because =85)
[rearranged]
Q deltaV /r =3D =BD kmv^2
=20
(1.6 X 10^ -19 C ) (4.5 X 10^3V) / 3,2 X10 ^-13 =3D =BD x 9.1 x 10^-31 X
v^2
v^2 =3D 4.94 X 10^24
v =3D &.03 X 10^13
What is r? Is not Q deltaV the work done by the electric field on the=20
electron? Other than that (and the "k" - assume a typo), it's a good=20
start.
If you introduced the r to try to find the electric field between the=20
plates, consider:
E =3D deltaV / r
F =3D E q
work done =3D F r
=3D E q r
=3D deltaV /r * q * r
=3D q deltaV
Part C comes into it because you used KE =3D 1/2 m v^2. This assumes that=
=20
the speed of the electron is a lot smaller than the speed of light. Is the=
=20
final speed of the electron large enough for KE =3D 1/2 m v^2 to be wrong?
5. Two parallel plates labeled W and X are separated by 5.2 cm. The
electric potential between the plates is 150V. An electron starts
from rest at time tW and reaches plate X at time tX. The electron
continues through the opening and reaches point P at time tp
(remember e =3D - 1.6 X 10^-19 C and the mass of an electron is 9.1 X
10^-31 kg)
=20
a.)=09Sketch the speed-time graph on the axes below
b.)=09Determine the kinetic energy of the electron as it arrives at
plate X=20
[rearranged]
FE =3D Ek
Q delta V/r =3D 1/3 mv^2
=20
1.6 x 10^-19 C (150V) / 0.05a =3D =BD (9.1X10^-31 kg) (v^2)
V=3D 2.9 X 10^-23 m/s
Same problem with the extra r as before.
--=20
Timo
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| User: "Uncle Al" |
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| Title: Re: hellllllllllllllllllllllllllllp..PLZ life or death situation |
11 Jan 2005 09:18:20 AM |
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hype_chicky wrote:
1. The electrostatic force between two point charges is 1.2 x 10^-2 N.
If the distance between them is doubled, the charge of one of the
point’s doubles and the charge of the other point triples. What will
be the force between them?
Kq1q2/r2
[snip]
Usenet's cloaca,
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Posted at:
www.GroupSrv.com
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Life or death? Then die. Free up resources for those who can make
use of them to society's benfit. Think of it as evolution in action.
Don't go into Sales or Marketing, either.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "Sam Wormley" |
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| Title: Re: hellllllllllllllllllllllllllllp..PLZ life or death situation |
10 Jan 2005 08:48:07 PM |
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hype_chicky wrote:
1. The electrostatic force between two point charges is 1.2 x 10^-2 N.
If the distance between them is doubled, the charge of one of the
points doubles and the charge of the other point triples. What will
be the force between them?
Kq1q2/r2
Read your text book
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