Understanding Electric Forces: Solutions to Group Problems 1
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University of Illinois, Chicago**We aren't endorsed by this school
Course
PHYS 142
Subject
Physics
Date
Dec 11, 2024
Pages
5
Uploaded by GrandSalamander4262
1 Physics 142 Group Problems 1 Solution This week’s new concepts: •Conductors allow charges to move until the net force on them is zero. •Force on charges: Coulomb’s Law says ࠵?!"=!#$%!|’"’#|("##o࠵?!"is the distance between charges ࠵?!and ࠵?"oLike charges repel each other, and opposite charges attract each other. oThe constant is ࠵?)=!#$%!= 8.99 × 10*+⋅-#.#, and ࠵?/= 8.85 × 100!".#+⋅-#. Problem 1:Electric forces and components A.A source charge exerts an electric force on a point charge of magnitude ࠵?!= 120 N, pointing 30°above the +࠵?axis (in the 1stquadrant). Find the ࠵?and ࠵?components of the force on this point charge. Since the force is in the 1stquadrant, both components will be positive: ࠵?!,2= (120 N) cos 30° = 104 N࠵?!,3= (120 N) sin 30° = 60 NB.Another source charge is added, creating a second force ࠵?"= (25 N)࠵?̂on the point charge. 1.Find the magnitude of the net force on the point charge. Add the components: ࠵?4)5,2= 104 N + 0 = 104 Nand ࠵?4)5,3= 60 N + 25 N = 85 N. Then use the Pythagorean theorem: ࠵?4)5= A(104 N)"+ (85 N)"= 134 N2.Suppose the magnitude of the point charge were doubled. How would this change the magnitude of the force you found above? Coulomb’s law says the electric force is proportional to both charges. Doubling one of them would double the force. This would double each of the applied forces, also doubling the net force. C.A third charge is added, so that the net force on the original point charge is zero. Find the components of the force due to the third charge. We know the net force is zero, so we can start by adding the three forces: ࠵?⃑4)5= 0 = ࠵?⃑!+ ࠵?⃑"+ ࠵?⃑6⇒ ࠵?⃑6= −࠵?⃑!− ࠵?⃑"This is true for the components as well as the entire force. In particular, the ࠵?-component of ࠵?⃑6must cancel those of ࠵?⃑!and ࠵?⃑", and likewise for ࠵?: ࠵?6,2= −104 N࠵?6,3= −85 N
2 Problem 2:Electric interactions A.Label each of the following statements as true or false. T Two positive charges will always repel each other, no matter how far apart they are. F The net electric force on a charge is only zero if there are no other charges around. B.Two students are discussing their answer to the 2ndstatement above. Student 1: “This must be true. Coulomb’s law says the force is proportional to ࠵?!࠵?", so it can only be zero if ࠵?"is zero.” Student 2: “I was thinking about the vector addition we did in problem 1. What if we had two forces which cancelled each other – wouldn’t this be false in that case?” With which student do you agree? Student 2 is correct. If the electric forces due to any other charges on one charge all cancel out, the net force would be zero even with other charges around.C.The diagram at right shows two point charges, +࠵?!on the ࠵?-axis at (0,2)and −࠵?"on the ࠵?-axis at (3,0). 1.A negative point charge −࠵?6is placed at the origin, (0,0). On the diagram, qualitatively sketch the force that each of the charges on the axes would exert on the charge at the origin, and then sketch the net force vector. The charge on the y-axis is the opposite sign so it will attract ࠵?6, and the charge on the x-axis is the same sign so it will repel. The net force must point somewhere in the 2ndquadrant. 2.Suppose the charge is replaced by a positive charge: ࠵?6,789= +࠵?6. How would the magnitude and direction of the net force on ࠵?6change? Since the force is proportional to the product of the charges, the magnitude of each force would not change but the direction of each would reverse. The net force would be the same magnitude and in the opposite direction. 3.Suppose instead that the charge is replaced by a charge that is twice as big, but with the same sign: ࠵?6,789= −2࠵?6. How would the net force on ࠵?6change? This would not change the direction but would double the magnitude instead. Suppose we adjust the charges on ࠵?!and ࠵?"so that the net force is at 45°from the ࠵?-axis (pointing due northwest). 4.For this to be true, would ࠵?"need to be larger than, smaller than, or equal to ࠵?!? For the net force to be at 45°, that means the force that each charge exerts on ࠵?6must be equal. Since ࠵?"is farther away from the origin than ࠵?!is, to exert the same force we must have ࠵?"> ࠵?!. q2q1q3112323
3 Problem 3:Coulomb’s law A.A charge ࠵?!= +1.5 nCis placed 40 cmfrom an unknown charge ࠵?". The charge ࠵?!experiences a repulsive force of 18.3 Nat that location. 1.What must the sign of ࠵?"be? Since the charges repel, they must be the same sign, so ࠵?"must be positive. 2.Find the charge ࠵?". We can rearrange Coulomb’s law to find the charge: ࠵? =14࠵?࠵?/࠵?࠵?࠵?"= ࠵?)࠵?࠵?࠵?"⇒࠵?࠵?"࠵?)࠵?= ࠵? = +0.22 C (positive since it repels ࠵?)The charge ࠵?!is moved to a distance of 1.0 mfrom ࠵?". 3.What would the force on ࠵?!be at this new location? We have the charges so we could plug in the values. We can also do this as a ratio: the distance is !.//.#= 2.5times farther away. Since the force is proportional to !(#, the force is !".;#= 0.16times as big, or ࠵?789= 0.16(18.3 N) = 2.9 NB.A third charge ࠵?6with half the value you found in part A is added to the system, so that the net force on ࠵?!is zero. The charge ࠵?"is still 40 cmfrom ࠵?!. 1.How far from ࠵?!must ࠵?6be to make the net force zero? Looking at Coulomb’s law for the second force, the numerator will be half as large. For the two forces to cancel, the denominator of the second force also has to be half as large: ࠵?6"=12࠵?""⇒ ࠵?6=1√2࠵?"= 28.3 cm2.Sketch a diagram showing the approximate locations of the three charges and their relative positions. The charges ࠵?"and ࠵?6must be on opposite sides of ࠵?!for the two repulsive forces to cancel, and as we found above ࠵?6must be closer: 3.Suppose the new charge ࠵?6had the same absolute value but was negative instead. Where would the charge have to be placed in this case, to make the net force zero? Sketch a diagram showing the approximate locations of the three charges and their relative positions. Now both ࠵?"and ࠵?6must be on the same side as ࠵?!(with the same distances we found) for the attractive and repulsive forces to balance: q1q2q3q1q2q3
4 Problem 4:Electric force components in 2D Three point charges are arranged on the corners of a rectangle as shown. The charges are ࠵?!= 2.0 µC, ࠵?"= 2.5 µC, and ࠵?6= 1.0 µC. A.First, sketch the electric forces acting on charge ࠵?!. B.Find the magnitude of the force on ࠵?!due to ࠵?". From Coulomb’s law: ࠵?!"= ࠵?)࠵?!࠵?"࠵?!""= S8.99 × 10*N ⋅ m"C"U(2.0 µC)(2.5 µC)(3.0 m)"= 4.99 × 1006NC.Now consider the force on ࠵?!due to ࠵?6. 1.Will this have positive or negative ࠵?̂and ࠵?̂components? It will have negative ࠵?̂and positive ࠵?̂components. 2.To find these components, we need to know sin ࠵?and cos ࠵?. Look at the triangle involving the lengths and ࠵?, and use that to find each trig function. (Hint: You do not need to find ࠵?first, but you will need the length of the hypotenuse.) As the hint suggests, first find the hypotenuse: √3"+ 5"= 5.8 mThen from the triangle, sin ࠵? =<==>?==6./;.@= 0.52and cos ࠵? =ABC>?==;./;.@= 0.86. 3.Find the magnitude of the force on ࠵?!due to ࠵?6. Now we have the hypotenuse, we can plug it into Coulomb’s law: ࠵?!6= S8.99 × 10*N ⋅ m"C"U(2.0 µC)(1.0 µC)(5.8 m)"= 5.3 × 100#N4.Use your answers above to find the horizontal and vertical components of this force. Multiplying the magnitude by the trig functions: ࠵?!6,2= −(5.3 × 100#N)(0.86) = −4.6 × 100#N࠵?!6,3= (5.3 × 100#N)(0.52) = 2.8 × 100#ND.Add the components of the forces you found to find the components of the net force on ࠵?!. ࠵?4)5,2= −4.6 × 100#N࠵?4)5,3= 4.99 × 1006N + 2.8 × 100#N = 5.27 × 1006NE.What is the magnitude of the net force? Using the Pythagoream theorem, ࠵?4)5= Y࠵?4)5,2"+ ࠵?4)5,3"= 5.29 × 1006Nq1q2q33.0 m5.0 mθ
5 Problem 5:Dipole A dipoleis a set of two equal and opposite point charges. It is characterized by the absolute value of the charges ࠵?and the distance between them ࠵?, as shown at right. A.A dipole is formed with ࠵? = 3.0 µCand ࠵? = 80 cm. What is the force that −࠵?exerts on +࠵?? The charges are opposite, so the −࠵?attracts +࠵?to the right. The magnitude is ࠵?)’#D#= [8.99 × 10*+⋅-#.#\(6 G.)#(/.@ -)#= 0.13 NB.A third charge ࠵?6= 1.5 nCis placed at point ࠵?, which is at the center of the dipole as shown. 1.Find the net force on ࠵?6. There are two forces on ࠵?6– a repulsive force to the right from the +࠵?point charge, as well as an attractive force to the right from the −࠵?point charge. Since point ࠵?is the same distance from each charge, and the charges have the same absolute value, each of these forces is the same: ࠵?4)5= ࠵?)|࠵? ∗ ࠵?6|[࠵?2\"+ ࠵?)|−࠵? ∗ ࠵?6|[࠵?2\"= 2࠵?)࠵? ∗ ࠵?6[࠵?2\"= 5.1 × 100#N2.The charge ࠵?6is now moved to point ࠵?, a distance ࠵?/4from the left point charge. Find the net force on ࠵?6at this location. This is the same expression as above, but with a different denominator: point ࠵?is 20 cmfrom +࠵?and 60 cmfrom −࠵?. ࠵?4)5= ࠵?)|࠵? ∗ ࠵?6|(20 cm)"+ ࠵?)|−࠵? ∗ ࠵?6|(60 cm)"= 1.1 × 1006NThis is about twice as large as the force at point ࠵?. +q–qd+q–qdP+q–qdd/4R