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[SOLVED] Electric Force
http://session.masteringphysics.com/problemAsset/1001926/10/knight_Figure_25_46.jpg
What is the force F on the 1 nC charge at the bottom?
Write your answer as two vector components, separated by a comma. Express each component numerically, in Newtons, to three significant figures.
F = kq1q2 / r^2
a^2 + b^2 = c^2
Sine Law
As a reference, the "focus" charge is the 1nC at the bottom.
Note that this appears to be a semicircle. Due to this fact, I conclude that the distance between the focus charge and the -6nC is also 5cm.
Because both 2nC charges have the same charge, distance, and angle with respect to the origin (Focus), I conclude that they cancel each other out along the X axis. They will both push the focus downwards, however there should be no X axis movement. Since the -6nC charge is along the Y axis, I conclude that this charge will not affect the focus' net X force either.
Therefore, F_x = 0
The force given between the 2nC on the left, and the focus, is F = kq1q2 / r^2
Where:
k = 9*10^9
q1 = 1*10^-9 (Focus)
q2 = 2*10^-9 (Leftmost charge)
r = 5cm = 0.05m
F = 0.0000072
Due to Sine Law, the X and Y components of this force are equivalent, since the angle is at 45 degrees.
Sqrt(X^2 + Y^2) = F^2
Sqrt(2Y^2) = F^2
2Y^2 = F^4
Y^2 = (F^4)/2
Y = Sqrt((F^4)/2)
Y = 3.6656415536710623664939771731515 * 10^-11
Since the rightmost charge is identical with respect to the focus, it will put out the same force along the Y axis.
Now to include the -6nC charge. F = kq1q2 / r^2
Where:
k = 9*10^9
q1 = 1*10^-9
q2 = -6*10^-9
r = 0.05
F = -0.0000216
So, the net force on the focus should be:
= "-6nC" + "2nC:Y" + "2nC:Y"
= "-6nC" + 2*"2nC:Y"
= -0.0000216 + 2*3.6656415536710623664939771731515 * 10^-11
= -0.00002159992668716892657875268
Rounded to 3 significant digits is 2.16*10^-5
Expressing answer as asked:
0,2.16*10^-5
This answer is incorrect.
My mistake is where?
Homework Statement
http://session.masteringphysics.com/problemAsset/1001926/10/knight_Figure_25_46.jpg
What is the force F on the 1 nC charge at the bottom?
Write your answer as two vector components, separated by a comma. Express each component numerically, in Newtons, to three significant figures.
Homework Equations
F = kq1q2 / r^2
a^2 + b^2 = c^2
Sine Law
The Attempt at a Solution
As a reference, the "focus" charge is the 1nC at the bottom.
Note that this appears to be a semicircle. Due to this fact, I conclude that the distance between the focus charge and the -6nC is also 5cm.
Because both 2nC charges have the same charge, distance, and angle with respect to the origin (Focus), I conclude that they cancel each other out along the X axis. They will both push the focus downwards, however there should be no X axis movement. Since the -6nC charge is along the Y axis, I conclude that this charge will not affect the focus' net X force either.
Therefore, F_x = 0
The force given between the 2nC on the left, and the focus, is F = kq1q2 / r^2
Where:
k = 9*10^9
q1 = 1*10^-9 (Focus)
q2 = 2*10^-9 (Leftmost charge)
r = 5cm = 0.05m
F = 0.0000072
Due to Sine Law, the X and Y components of this force are equivalent, since the angle is at 45 degrees.
Sqrt(X^2 + Y^2) = F^2
Sqrt(2Y^2) = F^2
2Y^2 = F^4
Y^2 = (F^4)/2
Y = Sqrt((F^4)/2)
Y = 3.6656415536710623664939771731515 * 10^-11
Since the rightmost charge is identical with respect to the focus, it will put out the same force along the Y axis.
Now to include the -6nC charge. F = kq1q2 / r^2
Where:
k = 9*10^9
q1 = 1*10^-9
q2 = -6*10^-9
r = 0.05
F = -0.0000216
So, the net force on the focus should be:
= "-6nC" + "2nC:Y" + "2nC:Y"
= "-6nC" + 2*"2nC:Y"
= -0.0000216 + 2*3.6656415536710623664939771731515 * 10^-11
= -0.00002159992668716892657875268
Rounded to 3 significant digits is 2.16*10^-5
Expressing answer as asked:
0,2.16*10^-5
This answer is incorrect.
My mistake is where?
Last edited: