Recent Changes

a) the object lies beyond the center of curvature
b) the object lies within the focal point
Given parameters
concave mirror
f = + 30 cm
a) If the object lies beyond the center of curvature, the image forms on the same side of the object, in front of the object.
Thus p - i = 45
i = p- 45
use the thin lens formula, 1/p + 1/i = 1/f
1/p + 1/(p-45) = 1/30
solve for p, you can either feed the information into the calculator or use the common denominator method - the basic math tricks
b) If the object lies within the focal length , the image forms behind the mirror (opposite to the object) - or forms a virtual image
Thus p + i = 45
i = 45 -p
use the thin lens formula, 1/p + 1/i = 1/f
1/p + 1/-(45-p) = 1/30
solve for p

02. Three point charges, each of magnitude 3 μC, are placed at the corners of an equi-lateral triangle 4 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive, find the magnitude of the force experienced by one charge at a corner. Answer in units of N
-- Please refer to the class notes. We discussed one similar to this !
Week 8 (26th, November 2012)
A concave mirror has a focal length of 30.0 cm. The distance between an object and its image is 45.0 cm. Find the object and image distances, assuming that,
a) the object lies beyond the center of curvature
b) the object lies within the focal point

--- Focus on exam ---
01. Four point charges, each of magnitude 2 μC, are placed at the corners of a square 10 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge.
AnswerYour knowledge in units of N.vectors, taking x and y components, finding net forces are extremely important in this section. You all must have done this in PHY 1 (it is a compulsory part in the course).
Please note, I can't re-teach this during the class time, and I am requesting you to refresh your knowledge and come to the class.
I am willing to help you outside the class, if necessary.
Steps to solve the problem
Draw a diagram
Mark the forces, considering the sign (+ or -) on the charges
Figure out (or calculate if necessary ) the distance between any two charges.
{Picture1-121-CF2.png}
{Picture4-121.png}
. {Wiki 121-CF.png}
02. Three point charges, each of magnitude 3 μC, are placed at the corners of an equi-lateral triangle 4 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive, find the magnitude of the force experienced by one charge at a corner. Answer in units of N
-- Please refer to the class notes. We discussed one similar to this !

A nice video for you to go through the Coulomb's Law
--- Focus on exam ---
negative charge. Answer
Answer in units of N. {Wiki 121-CF.png}
02. Three point charges, each of magnitude 3 μC, are placed at the corners of an equi-lateral triangle 4 cm on a side. The value of Coulomb’s constant is 8.98755 × 10^9 N m²/C². If three of the charges are positive, find the magnitude of the force experienced by one charge at a corner. Answer in units of N

1. I will call q1 the negative charge in which we are trying to find the force acted on it. Q2,q3,q4 are all positive. This means they all have attractive forces to the q1. I set the charges up so that q1 is in the bottom left corner, q2 is the upper left corner, q3 is the upper right corner, and q4 is the lower right corner. We know:
q=2 microC
k= is a constant so we don't have to substitute an actual numberk=8.9 e9
r= the 10 cm distance between charges which is 0.1 m
Based on this information we can use the equation f=k ((q1)(q2))/ r^2, to find the force acting on q1 due to q2 and q4.
F12= k(2microC)(2microC)/(0.1^2) = 0.4 nN,k(2microC)(2microC)/(0.1^2), this also
To find F13 we need to find the difference between q1 and q3 since they are diagonal from each other.
in: a=(square root)(0.1m^2)+(0.1m^2)= 0.14 mroot)(0.1m^2)+(0.1m^2)
Now we for F13=k(2microC)(2microC)/(0.14^2) = 0.204 nN
Next we have to find the sum of the forces in the x and y directions.
The sum of forces in the x direction = f13 cos60+f14cos0
q=3 microC
r=4cm=0.04m
k=constantk=8.9 e 9
Choosing q1 to be the test charge:
F12=k q1 q2/r^2
Find the sum of the forces in the y direction = F12
F net= (square root) (sum of x^2)+(sum of y ^2)

Do the algebra and solve for L
L=2.12 m
Solutions: Week 7:
1. I will call q1 the negative charge in which we are trying to find the force acted on it. Q2,q3,q4 are all positive. This means they all have attractive forces to the q1. I set the charges up so that q1 is in the bottom left corner, q2 is the upper left corner, q3 is the upper right corner, and q4 is the lower right corner. We know:
q=2 microC
k= is a constant so we don't have to substitute an actual number
r= the 10 cm distance between charges which is 0.1 m
Based on this information we can use the equation f=k ((q1)(q2))/ r^2, to find the force acting on q1 due to q2 and q4.
F12= k(2microC)(2microC)/(0.1^2) = 0.4 nN, this also equals f14 because all of the numbers are the same.
To find F13 we need to find the difference between q1 and q3 since they are diagonal from each other.
We use the equation a^2=b^2+c^2, since b and c are both 10 cm we can plug in: a=(square root)(0.1m^2)+(0.1m^2)= 0.14 m
Now we can plug in for F13=k(2microC)(2microC)/(0.14^2) = 0.204 nN
Next we have to find the sum of the forces in the x and y directions.
The sum of forces in the x direction = f13 cos60+f14cos0
The sum of the forces in the y direction = f13sin60+f12cos0
Plug in and get two answers. In order to find the net force we have to square those two numbers and then square root them.
Fnet= (square root)(sum of fx^2+sum of fy^2)
2. q1 is placed on the bottom left, q2 is the top of the triangle, and q3 is the bottom right corner (just what i chose for my diagram). Since all charges are positive, they are repulsive to each other. We know:
q=3 microC
r=4cm=0.04m
k=constant
Choosing q1 to be the test charge:
F12=k q1 q2/r^2
F13=k q1 q2/r^2
Plug in and both answers will be the same since all of the givens are the same.
Find the sum of the forces in the x direction = F13
Find the sum of the forces in the y direction = F12
F net= (square root) (sum of x^2)+(sum of y ^2)

--> V = 340m/s
--> use L from first step
Third Step:[[#|Step]]:
Frequency of 2nd Harmonic of Pipe A = Frequency of Third Harmonic of Pipe B
fB = nV/4L ----> rearrange to find L
Now 11.975ohm is in parallel with 20ohm --> 1/11.975 + 1/20 = 1/0.1335 = 7.49ohm
Net Resistance = 7.49ohm
find Current:
V = IR ---> Given: V=6V; R = Net resistance from part 1
= V/R
I = (6V)/(7.49ohm)
I = 0.801A
__
Week 7
1.
Four point charges in a square (10cm a side)
--> Q1 at bottom left corner (-2uC)
--> Q2 at top left corner (+2uc)
--> Q3 at top right corner (+2uc)
--> Q4 at bottom right corner (+2uc)
*Q1 and Q3 are not directly touching within the square... they act on each other through the center of the square; cutting the square diagonally in half (in two triangles)
-----> to find hypotenuse: a^2 + b^2 = C^2 (a is side 1 of triangle, b is side 2, c is hypotenuse)
----> 10^2 +10^2 = c^2 => c = 14.14cm --> this is the distance to find F13
*1uC = 10^-9C
Use eqnt: F = (kQaQb)/r^2 for each force acting on Q1:
F14 = (kQ1Q4)/r^2 --> F14 = [(8.98755e9)(2e-9C)(2e-9C)]/(0.10m^2) ---> F14 = 3.59e-6N
F12 = (kQ1Q2)/r^2 --> F12 = 3.59e-6N
F13 = (kQ1Q2)/r^2 --> *Be sure to use 14.14cm instead of 10cm* --> F13 = 1.79e-6N
Now Find the X and Y Components:
SumFx = F14cos(0) + F13cos(45) + F12cos(90) ---> Q14 is directly on x axis, thus why cos(0); Q13 is between 0 and 90, thus cos(45); Q12 is 90 degrees from x axis thus cos(90)
SumFx = 4.85e-6N
SumFy = F14sin(0) + F13sin(45) + F12sin(90)
SumFy = 4.85e-6N
To find Fnet:
Fnet = sqrt(Fx^2 + Fy^2)
Fnet = 6.86e-6N
2.
Three point charges in corners of an equilateral triangle (4cm each side)
-->Q1 in bottom right corner (+3uC)
--> Q2 top middle (+3uC)
--> Q3 bottom left corner (+3uC)
Use eqn: F = (kQaQb)/r^2 for each force acting on Q1
F12 = (kQ1Q2)/r^2 --> F12 = 5.06e-5N
F13 = (kQ1Q3)/r^2 --> 5.06e-5N
Now find X and Y Components:
SumFx = F13cos(0) + F12cos(60) ---> equilateral trial has all three angles at 60degrees
SumFx = 7.59e-5N
SumFy = F13sin(0) + F12sin(60)
SumFy = 4.38e-5N
To find Fnet:
Fnet = sqrt(Fx^2 +Fy^2)
Fnet = 8.76e-5N