16_1 Quiz Help
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Re: 16_1 Quiz Help
Number three is just a plugin question basically.
For number four it helps to draw a line. Find the potential for the charge you are given on at the position they are asking for. Then, since the question asks for a net potential of zero, you know that the potential you are looking for has to be equal in magnitude and opposite in sign to the one you first found. Just use the formula V=kQ/r. They have the same potential and k constant but you will have a different distance, just plug in and solve for Q.
Any ideas for number 2? I know there are four charges and that you have to add up the potentials, but I'm not sure where to go from there.
For number four it helps to draw a line. Find the potential for the charge you are given on at the position they are asking for. Then, since the question asks for a net potential of zero, you know that the potential you are looking for has to be equal in magnitude and opposite in sign to the one you first found. Just use the formula V=kQ/r. They have the same potential and k constant but you will have a different distance, just plug in and solve for Q.
Any ideas for number 2? I know there are four charges and that you have to add up the potentials, but I'm not sure where to go from there.
Andrew Guest
Re: 16_1 Quiz Help
For number 2:
You need to multiply the charge by 4. Your r will be the distance to the center of the square. Then you just plug it into 16.1
You need to multiply the charge by 4. Your r will be the distance to the center of the square. Then you just plug it into 16.1
c Guest
question 2,3,4
i've tried these questions so many times but still can't get them please HELP!! =/ thanksss!
ccc Guest
question 2
the above post for number two is correct. i just want to clarify how to figure out the true r to use.
Think about a square, and a line running through the center connecting two of the corners. You get two right triangles, and the center charge falls on the line. If the charge is on the dead center of that line, (the hypotenuse of the triangle) how can you figure out the distance that point is from all other the other charges?
(start by figuring out the length of the hypotenuse)
cheers.
Think about a square, and a line running through the center connecting two of the corners. You get two right triangles, and the center charge falls on the line. If the charge is on the dead center of that line, (the hypotenuse of the triangle) how can you figure out the distance that point is from all other the other charges?
(start by figuring out the length of the hypotenuse)
cheers.
omghi2u Guest
ch16_1 #2
so is this the right set up for #2?? cuz i keep gettin it wrong and its frustrating me!!
the equation I'm getting is: PE (the answer) = (9.0e9) ((incoming charge)(4*charges on the corners)/(side of square/2))
whats weird is when you use Pythagorean Theorem and divide that answer by 2 for the midpoint, you get a different number for r, which doesnt make sense  so which way is the right way?
where am I going wrong? am i overthinking this?
the equation I'm getting is: PE (the answer) = (9.0e9) ((incoming charge)(4*charges on the corners)/(side of square/2))
whats weird is when you use Pythagorean Theorem and divide that answer by 2 for the midpoint, you get a different number for r, which doesnt make sense  so which way is the right way?
where am I going wrong? am i overthinking this?
confuzed Guest
#4
Does anyone know how to do number 4 I have spent a while on it and I cant figure it out?
sjames Guest
2 and 4
Okay for 2 I used the equation
V=k*Q/r
to get r I took the length L in cm that was given then I divided it by 2 then by cos 45. Kind of like on p.8 for ch. 16, they did 60 b/c it was a triangle we are using 45 because it is a square.
so
(L/2)/cos45=r
then I multiplied it by 4 b/c V=V_1+V_2+V_3+V_4, remember all sides need to be taken to get the magnitude
then multiply by the q_0 you got. My answer was +26 to give you some idea
remember SI, stuff
4
look at p.11 in ch.16 it's really just plug and chug. It should be negative
V=k*Q/r
to get r I took the length L in cm that was given then I divided it by 2 then by cos 45. Kind of like on p.8 for ch. 16, they did 60 b/c it was a triangle we are using 45 because it is a square.
so
(L/2)/cos45=r
then I multiplied it by 4 b/c V=V_1+V_2+V_3+V_4, remember all sides need to be taken to get the magnitude
then multiply by the q_0 you got. My answer was +26 to give you some idea
remember SI, stuff
4
look at p.11 in ch.16 it's really just plug and chug. It should be negative
student Guest
q4
I tried to plug and chug for number 4 it doesn't I think I am doing something wrong but I don't know what can someone explain plz?
sjames Guest
question 2
hey, i tried every method listed and i'm getting a realllyyyy big number 25377679468 o.o
i converted the corner charges (q_0) and magintude charge (Q) to Coloumbs by multiplying by 10E06
the distance to m by dividing by 100
then for the equation V = (k*Q)/(r)
to get r: i took the side of the rectangle given and divided it by 2. then i also divided it by cos(45)
and plugged it into the equation with the Q being the larger given charge and k being 9E9
then i multiplied this by 4 because there are 4 charges for the corners
then i multipled this number by q_0, the smaller individual corner charge for the equation
W= q_0 * V
i still am not getting this. please help if you can i've been stuck on this problem for ages o.o
i converted the corner charges (q_0) and magintude charge (Q) to Coloumbs by multiplying by 10E06
the distance to m by dividing by 100
then for the equation V = (k*Q)/(r)
to get r: i took the side of the rectangle given and divided it by 2. then i also divided it by cos(45)
and plugged it into the equation with the Q being the larger given charge and k being 9E9
then i multiplied this by 4 because there are 4 charges for the corners
then i multipled this number by q_0, the smaller individual corner charge for the equation
W= q_0 * V
i still am not getting this. please help if you can i've been stuck on this problem for ages o.o
:) Guest
Question 3 in Detail
The magnitude of the potential at a distance of 18.76 cm from a source charge is 5.166 kV. What is the magnitude of the source charge (see sheet 5) ? Indicate with a positive (negative) sign whether the electric field vector is tangential (perpendicular) to the equipotential lines for the source charge (see sheet 12 ).
How To Solve
I think the first and easiest step is to make sure we're in S.I. Units, which in this case we are not. So let's convert! Distance needs to be in meters (m) and the charge needs to be in Volts (V)
18.76 cm = 0.1876 m
5.166 kV (that's kilivolts) = 5166 V
Remember the prefix kili converts the same for any unit which is a thousand.
If we reference CH 16 Sheet 12 we will find two formulas we can use. Since we already have a charge and the distance we can use Equation 16.3 which is:
Where:
V is the volts (which we are given)
r is the distance [technically radius] between the charge (which we also have)
k is a constant which never changes and the number is 9 x 10^{9} which I will simpliy as "9E9".
All that's left is Q the charge which the question is asking for and we can now figure out.
Sometimes it's easier for me to substitute all the numbers and then proceed to solve using algebra and real numbers then just doing algebra to find a new formula, so I will do it that way.
Using
V = k*(Q/r)
5166 = 9E9*(Q/0.1876)
To get rid of the fraction I multiply both sides by the denominator.
[5166*0.1876] = 9E9* [(Q/0.1876)*0.1876]
969.1416 = 9E9*Q
Now we want Q to be all by itself since that is the variable we want. In order to do this (for this example) I will divide out the constant "9E9", and what I do to one side I must do to the other.
[969.1416/9E9] = [(9E9*Q)/9E9]
The "9E9" on the right side cancel out
969.1416/9E9 = Q
And this will give our final answer
I am having little luck with Question 4, I know there are two parts to it but I can't seem to figure out the first part.
Also does anyone know if the Assignment is due Thursday night at midnight? Or at midnight Thursday morning (which at the time of this post has already past)?
How To Solve
I think the first and easiest step is to make sure we're in S.I. Units, which in this case we are not. So let's convert! Distance needs to be in meters (m) and the charge needs to be in Volts (V)
18.76 cm = 0.1876 m
5.166 kV (that's kilivolts) = 5166 V
Remember the prefix kili converts the same for any unit which is a thousand.
If we reference CH 16 Sheet 12 we will find two formulas we can use. Since we already have a charge and the distance we can use Equation 16.3 which is:
V = k*(Q/r)
Where:
V is the volts (which we are given)
r is the distance [technically radius] between the charge (which we also have)
k is a constant which never changes and the number is 9 x 10^{9} which I will simpliy as "9E9".
All that's left is Q the charge which the question is asking for and we can now figure out.
Sometimes it's easier for me to substitute all the numbers and then proceed to solve using algebra and real numbers then just doing algebra to find a new formula, so I will do it that way.
Using
V = k*(Q/r)
5166 = 9E9*(Q/0.1876)
To get rid of the fraction I multiply both sides by the denominator.
[5166*0.1876] = 9E9* [(Q/0.1876)*0.1876]
969.1416 = 9E9*Q
Now we want Q to be all by itself since that is the variable we want. In order to do this (for this example) I will divide out the constant "9E9", and what I do to one side I must do to the other.
[969.1416/9E9] = [(9E9*Q)/9E9]
The "9E9" on the right side cancel out
969.1416/9E9 = Q
And this will give our final answer
 Spoiler:
 1.076824E7
I am having little luck with Question 4, I know there are two parts to it but I can't seem to figure out the first part.
Also does anyone know if the Assignment is due Thursday night at midnight? Or at midnight Thursday morning (which at the time of this post has already past)?
Guest01 Posts : 133
Join date : 20080919
Re: 16_1 Quiz Help
I assumed that they meant thursday night midnight... at least i hope now because i didnt even think about it like that!
Kathleen Posts : 42
Join date : 20081218
Re: 16_1 Quiz Help
Hey Smiley Face,
Are you converting to 10E6 or 10E6? That may b your problem.
Are you converting to 10E6 or 10E6? That may b your problem.
Guest01 Posts : 133
Join date : 20080919
Question 4: Finally!
You really have to sit down with this one. Some trial and error on my part, but when I really sat down, drew a diagram and interpreted what the question was really asking it becomes much easier to see where one could deviate from the correct answer.
Two charges are placed on the xaxis, an unknown charge at = 5 cm and a positive 2.202 µC charge at 13.92 cm from the origin. What is the charge at = 5 cm which causes the total potential due to both charges at 8.549 cm to be zero, including the sign (see sheet 5,11) ?
First thing I always find easy and to get started on a problem... Convert to S.I.! In total there are three points which we will call x_{1}, x_{2}, and x_{3}.
x_{1} = 5 cm => 0.05 m
x_{2} = 13.92 cm => 0.1392 m
x_{3} = 8.549 cm => 0.08549 m
Q_{1} = 2.202 µC => 2.202E6
Q_{2} = ?????
Let's try to understand the question now and represent it in a drawing. This is ironic to me because I originally wanted to be an artist, so here goes my artistic skills as per the computer:
We have a line:
_______________________________________________________
And on this line we have two charges, the first is at 0.05 m, the second, opposite charge is at 0.1392
______x_{1}__________________________________________________________x_{2}
Now in between these two points there is a point where both charges cancel. This is measured as the third point, 0.08549.. What we are overlooking is that this is our origin and the radius goes out from here. We now have to calculate our two radii.
______x_{1}_____________________x_{3}__________________________x_{2}
So our origin goes from the central point x_{3} out towards both x_{2} and x_{1}, which we will label r_{1} and r_{2}, respectively.
r_{1} = x_{3}  x_{2}
r_{2} = x_{3}  x_{1}
To save some time, you should wind up with these values.
r_{1} = 0.05371
r_{2} = 0.03549
We can now find the PE for one charge (the charge that's given) and then use that to find the other charge.
The formula for PE (slightly modified for this question)
Where as you should get...
PE = 368981.5677
I am going to set this equal to V which is nearly the same formula. To understand how it will all work out I'm writing out the formula like this:
V = k*(Q_{2}/r_{2})[/quote]
368981.5677 = 9E9*(Q_{2}/0.03549
Work it out as you did in Question 3 and you should get around
Two charges are placed on the xaxis, an unknown charge at = 5 cm and a positive 2.202 µC charge at 13.92 cm from the origin. What is the charge at = 5 cm which causes the total potential due to both charges at 8.549 cm to be zero, including the sign (see sheet 5,11) ?
First thing I always find easy and to get started on a problem... Convert to S.I.! In total there are three points which we will call x_{1}, x_{2}, and x_{3}.
x_{1} = 5 cm => 0.05 m
x_{2} = 13.92 cm => 0.1392 m
x_{3} = 8.549 cm => 0.08549 m
Q_{1} = 2.202 µC => 2.202E6
Q_{2} = ?????
Let's try to understand the question now and represent it in a drawing. This is ironic to me because I originally wanted to be an artist, so here goes my artistic skills as per the computer:
We have a line:
_______________________________________________________
And on this line we have two charges, the first is at 0.05 m, the second, opposite charge is at 0.1392
______x_{1}__________________________________________________________x_{2}
Now in between these two points there is a point where both charges cancel. This is measured as the third point, 0.08549.. What we are overlooking is that this is our origin and the radius goes out from here. We now have to calculate our two radii.
______x_{1}_____________________x_{3}__________________________x_{2}
So our origin goes from the central point x_{3} out towards both x_{2} and x_{1}, which we will label r_{1} and r_{2}, respectively.
r_{1} = x_{3}  x_{2}
r_{2} = x_{3}  x_{1}
To save some time, you should wind up with these values.
r_{1} = 0.05371
r_{2} = 0.03549
We can now find the PE for one charge (the charge that's given) and then use that to find the other charge.
The formula for PE (slightly modified for this question)
PE = 9e9*(2.2026/0.05371)PE = k*(Q_{1}/r_{1})
Where as you should get...
PE = 368981.5677
I am going to set this equal to V which is nearly the same formula. To understand how it will all work out I'm writing out the formula like this:
. What we are looking for is Q_{2}, and because the formula is essentially the same I can work it backwards and use the same algebra we've been using.V = k*(Q_{2}/r_{2})
V = k*(Q_{2}/r_{2})[/quote]
368981.5677 = 9E9*(Q_{2}/0.03549
Work it out as you did in Question 3 and you should get around
 Spoiler:
 1.45502E6
Guest01 Posts : 133
Join date : 20080919
Re: 16_1 Quiz Help
Wow thanks so much Guest01 for the detailed explanations. I was just making a stupid mistake on #3 which if I realized earlier I would've figured it out hours ago but your answer helped me catch that. And i really appreciate the drawing out of number 4, I was having trouble understanding the concept behind the drawing and what not but it's really not that bad if you draw it out right. =]
Kathleen Posts : 42
Join date : 20081218
Question 1 16_1
6.243 = (9E9)*(Q*1.579E6/.05967)
6.9355E10 = (Q*1.579E6/.05967)
how did you get 6.9355E10?
6.9355E10 = (Q*1.579E6/.05967)
how did you get 6.9355E10?
Guest1 Guest
Re: 16_1 Quiz Help
i still get 3 and 4 wrong
even though i followed the step and worked out the ones posted here
even though i followed the step and worked out the ones posted here
mia Guest
question2?
Ive tried question 2 three or four times and havent gotten it right.. annd i cant completely figure out the explainations on the forum can someone help?
guest123 Guest
QUESTION 14
#1:
PE=k(Q*q/r)
#2:
r=L/2/cos45
v1=kq/r
v1*4*#2charge
#3:
answer is 
v=kq/r
#4:
answer is 
v=(k(+Q)/r)+(k(Q)/r)=0
PE=k(Q*q/r)
#2:
r=L/2/cos45
v1=kq/r
v1*4*#2charge
#3:
answer is 
v=kq/r
#4:
answer is 
v=(k(+Q)/r)+(k(Q)/r)=0
jsyz Posts : 7
Join date : 20081124
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