Electric Charge
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
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