A point charge q 1 = -2.3 C is located at the origin of a co-ordinate system. An
ID: 1335361 • Letter: A
Question
A point charge q1 = -2.3 C is located at the origin of a co-ordinate system. Another point charge q2 = 6.3 C is located along the x-axis at a distance x2 = 6 cm from q1.
1)
What is F12,x, the value of the x-component of the force that q1 exerts on q2?N
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Your submissions:
-36.225
Computed value:
-36.225
Submitted:
Wednesday, October 7 at 7:02 PM
Feedback:
Your answer has been judged correct; the exact answer is: -36.225.
2)
Charge q2 is now displaced a distance y2 = 2.2 cm in the positive y-direction. What is the new value for the x-component of the force that q1 exerts on q2?
N
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Your submissions:
-31.93
Computed value:
-31.93
Submitted:
Wednesday, October 7 at 10:40 PM
Feedback:
Be careful. I think you are on the right track, but remember we want the x-component of F12, not the magnitude of F12.
3)
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net= 10.644 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?
C
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4)
How would you change q1 (keeping q2 and q3 fixed) in order to make the net force on q2 equal to zero?
Increase its magnitude and change its sign
Decrease its magnitude and change its sign
Increase its magnitude and keep its sign the same
Decrease its magnitude and keep its sign the same
There is no change you can make to q1 that will result in the fet force on q2 being equal to zero.
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5)
How would you change q3 (keeping q1 and q2 fixed) in order to make the net force on q2 equal to zero?
Increase its magnitude and change its sign
Decrease its magnitude and change its sign
Increase its magnitude and keep its sign the same
Decrease its magnitude and keep its sign the same
There is no change you can make to q3 that will result in the fet force on q2 being equal to zero.
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(Survey Question)
6)
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Explanation / Answer
1)
q1 = -2.3 C = -2.3*10^-6 C
q2 = 6.3 C = 6.3*10^-6 C
F = k *q1 *q2 / r^2
=9*10^9*-2.3*10^-6*6.3*10^-6 / 0.06^2
= -36.225 N
========================================
2)
y2 = 2.2 cm
x2 = 6 cm
Theta = tan^-1(y2 / x2)
=tan^-1(2.2 / 6) + 180 degrees
=20.136 + 180
=200.136 degrees
r^2 = (x2)^2 + (y2)^2
=0.06^2 + 0.022^2
= 0.0041 m^2
F(x component) = k *q1 *q2*cos theta / r^2
=9*10^9*-2.3*10^-6*6.3*10^-6*cos 200.1 / 0.0041
= -29.87 N
=================================================
3)
halfway between q1 and q2 is
r = 0.5 *sqrt (0.06^2+0.022^2)
= 0.032 m
Electric field = 9*10^9* (2.3*10^-6 + 6.3*10^--6) / (0.032^2)
= 7.56*10^7 J/C ( positive towards the negaitve or to the left, towards the origin )
q3 = 10.644 / 7.56*10^7
= -0.141*10^-6 C
q3 is negative because it is pointing towards the positve
and attracted to the positive as it is repulsed from the negative,
=================================================
4)
Increase its magnitude and change its sign
by adding together forces of q1 and Fq3 gives a positive value so we need to increase the negative value q1.
===============================================================
5)
Decrease its magnitude and keep its sign the same
we can decrease the positive value by decreasing the positive q3 charge
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