# Electric Charge

Pages: 5 (1344 words) Published: July 9, 2013
HW 1 solutions Point Charge in One Dimension

A point charge q1 = -3.5 μC is located at the origin of a co-ordinate system. Another point charge q2 = 5.1 μC is located along the x-axis at a distance x2 = 9.3 cm from q1. 1) What is F12,x, the value of the x-component of the force that q1 exerts on q2? -18.57

N

For all of these problems we want to make use of the standard electric force equation: ̂ So for this problem with K=9*109 Nm2/C2, Q1=-3.5μC, Q2=5.1 μC, and r=9.3 cm we get F=-18.57 N. It’s important to realize that getting a negative force value means the charges attract. The direction of that attraction is determined by which charge we’re looking at.

2) Charge q2 is now displaced a distance y2 = 2.7 cm in the positive y-direction. What is the new value for the xcomponent of the force that q1 exerts on q2? -16.45

N

Same idea as 1 but with a different r. We use the Pythagorean theorem to find the new distance which is 9.684 cm. Using the force equation again we get F=-16.45

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 = 5.71 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3? 1.167

μC

Here since all the charges are on a line we don’t have to worry about different directions. We had to calculate the force on Q2 do to Q1 for our earlier problems which was -17.13. Since we know the total force we can just say -17.13+F3=5.71. That implies that F3=22.84 N. Using the electronic force equation and knowing the k and the charge of Q2 we can find Q3 if we can find the distance between charge Q3 and Q2. Since we know r12=2r23 then the distance we’re looking for is 4.842 cm. Running that through the equation we get Q3=1.1667 μC. Point Charge in Two Dimensions

Three charges (q1 = 5.8 μC, q2 = -5.9 μC, and q3 = 3.4μC) are located at the vertices of an equilateral triangle with side d = 9.3 cm as shown. 1) What is F3,x, the value of the x-component of the net force on q3? 20.697

N

We can use superposition here to find the force. That means that when finding the force on a charge we can look at the contributions of the other charges individually and then add them all

together. Once again we use the electric force equation and find the force between each of the charges. We have: F12=-35.60874 N F13=20.520291 N F23=-20.874 To find the Fx on the third charge we take the appropriate forces and use trigonometry to get the x-components. Then Fx=(F13+F23)/2=20.697 to the right. The attractive force between 2 and 3 adds to the pushing force from 1 on 3. 2) What is F3,y, the value of the y-component of the net force on q3? -.306

N

Following the same kind of logic we get Fy=( F13+F23)* =-0.306 N

A charge q4 = 3.4 μC is now added as shown. What is F2,x, the x-component of the new net force on q2? -56.48

N

We repeat the steps above but we need to find the forces due to the new particle. Since it’s charge is the same as that due to Q3 then F14=20.520 N F24=-20.874 N And F2X= F12+( F24+ F23)/2=-56.48 N i.e. to the left

4) What is F2,y, the y-component of the new net force on q2? 0

N

Since we’ve added a charge opposite Q3 that’s exactly the same charge and distance away it cancels out its effect and the Fy on Q2 just becomes zero since Q1 has no Y component 5) What is F1,x, the x-component of the new net force on q1? 15.089

N

. Using the forces we calculated above and employing the same procedure we get F1X=F12-(F14+F13)/2=15.09 N. it’s important to note Q2 pulls to the right but Q3 and Q4 push to the left.

Electric Field from Point Charges

Two point charges (q1 = -4.2μC and q2 = 6.1 μC) are fixed along the x-axis, separated by a distance d = 9.3 cm. Point P is located at (x,y) = (d,d). 1) What is Ex(P), the value of the x-component of the electric field produced by q1 and q2 at point P? -1545190

N/C

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For this one we just use the equation:...