15.3 #4 HELLPPPPPPPPP

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15.3 #4 HELLPPPPPPPPP

Post  ahhhhhhh on Fri Feb 27, 2009 6:08 pm

has anyone tried 15.3 ? im stuck on number 4....

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Re: 15.3 #4 HELLPPPPPPPPP

Post  teachmep on Fri Feb 27, 2009 8:22 pm

CH15_3 HELP!!!! pleeeeeease..and thank you?

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No. 4

Post  Lali on Sat Feb 28, 2009 4:38 pm

Suppose that the percentage they gave u is 30%.

You know that the max value is 180 degrees and cos(180) is -1 and if we decrease by 30% it will be -1*.3=-.3
Therefore, if we decrease by -.3 it is -1-(-.3)=-.7 and then u just take the inverse cos of this number to find the angle.

Anyone, number 2? I've tried everything with no. 2. Thanks.

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question 4

Post  shouldbe on Sat Feb 28, 2009 4:58 pm

Question 4:

Start off with defining when PE is max and min for a dipole.

If you play with the equation for PE for a dipole:

PE = U = -pEcosΘ

(Make sure you're in degrees)

You should figure out that when Θ = 180 you have Max PE and when Θ = 0 you have Min PE.
( cos(0) = 1 ; cos(180) = -1 ) *don't forget negative sign in equation!)

Also, remember that PE min occurs when the + end of the dipole is perfectly aligned and closer to the direction of the electric field (closest to the direction a negative charge, somewhere).
Likewise, Max PE occurs when the dipoles are reversed, (when the + charge is perfectly aligned with the E. Field. furthest away from the arrow-head)

So, if you lose some percent (XX.XX%) of your PE, that means you still have 1-.XXXX of your original (max) PE.

And since, ( cos(180) = -1 ) ; cos-1(1-.XXXX) = The angle which your dipole has moved. And since we always measure Θ with respect to the + pole and the overall direction of the E. Field. Take [ 180 - the angle you got ].

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Re: 15.3 #4 HELLPPPPPPPPP

Post  guest_01 on Sat Feb 28, 2009 5:09 pm

can anyone help out with #1 and #2 thanks!

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question 2

Post  shouldbe on Sat Feb 28, 2009 5:12 pm

st, for the love of god, make a picture.

If you do that, you'll figure out that there is definitely an eclectic field.

Where E = kQ/(r^2)

First things first, let's get r.

Since we have a square, and our point of interest is the center of the square. Use a^2 + b^2 = c^2 to get the length of the diagonal. Half that distance is the location of the center point, for all four corners of the square.

Next, figure out an Electric Field for one of the corners, pick one, it doesn't matter. We'll call it 'X'

You should get some large number, don't worry. Next, recall that your only care about the direction of this Electric Field influencing a point in the center, since a square is made up of four right angles, and the point you care about is in the dead center, your diagonal (hypotenuse) where the point lies makes a 45 degree angle with all of the corners.

Since you got the electric field magnitude for one corner, give it a direction now. Which direction? A 45 degree angle.

So take X and multiply it by sin45, make sure you're in degrees you fool.

Great, now you've got your E. Field with a direction, woot.

And I'll leave it to you to figure out how many charges influence the field in this square. Multiply that number by the E.Field(sin45).....(its 4.)

All the charges are the same, therefore the fields they produce are all the same. Cheers.

It's negative because if all the charges were positive, the field would completely cancel itself out. (Draw arrows)

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q1

Post  student on Sat Feb 28, 2009 9:19 pm

Has anyone gotten q1 yet?

I was originally thinking that to get the E field to be 0 you need to cancel out the charges so I thought the positive charge ought to be the same number as the negative charge. But this doesn't work.
Is it just a matter of using the fact that electron charges are 1.6*10^-19?

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q 1

Post  student on Sat Feb 28, 2009 9:44 pm

Neverming I got it.
It goes E=k*Q/r^2 In this case you have distances which I will call X_1 (located at 0cm) X_2 (located further away) X_3 (furthest point away and the E field that is supposed to be 0)

Q(charge unkown)*k/ (X_3-X_1)^2 = Q (given) *k/(X_3-X_2)^2
the answer should be negative

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Re: 15.3 #4 HELLPPPPPPPPP

Post  student on Wed Mar 04, 2009 10:00 pm

I still cant get #2 following shouldbe's explanation.
I got E which is simple enough then i multiplied that by sin45 and then by 4
is this right? or did miss something.

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Re: 15.3 #4 HELLPPPPPPPPP

Post  shouldbe on Wed Mar 04, 2009 11:40 pm

did you half your hypotenuse?

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Re: 15.3 #4 HELLPPPPPPPPP

Post  student on Thu Mar 05, 2009 2:52 am

yea i halved it after using the original cube side given plugging it into a^2+b^2=c^2
I also tried just halving the side in the beginning and then using pythagorean but still doesnt work.

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iM STILL having problems with 3

Post  DELILA on Thu Mar 05, 2009 1:34 pm

Question 3: (1 point)

The electric dipole moment of a dipole with a charge magnitude of 0.2023 x10-19C is 2.228 x10-30Cm. How far are the charges apart (see sheet 27,30) ? Indicate with a positive (negative) sign whether the charges are equal or opposite in sign.

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iM STILL having problems with 3

Post  DELILA on Thu Mar 05, 2009 1:43 pm

Question 3: (1 point)

The electric dipole moment of a dipole with a charge magnitude of 0.2023 x10-19C is 2.228 x10-30Cm. How far are the charges apart (see sheet 27,30) ? Indicate with a positive (negative) sign whether the charges are equal or opposite in sign.

DELILA
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Re: 15.3 #4 HELLPPPPPPPPP

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