Topic 15: Electric Forces and Field(15.1 - 15.6)

Problem 5.2: Force on a Charge Three charged particles(1, 2, and 3) are used in this problem: q1 = 1.6 X 10^-19 C q2 = 3.2 X 10^-19 C q3 = -3.2 X 10^-19 C (a) Suppose charges 1 and 2 are placed alone on an x-axis and separated by a distance R = 0.0200 m, What is the force on 1 by 2? (b) Suppose charge 3 is placed directly between 1 and 2(on the x-axis) at a distance 3/4 R from particle 1, What is the force on 1? (c) Suppose charge 3 is 3/4 R from particle 1, but on a line that makes an angle θ = 60° with respect to the x-axis, What is the force on 1?

For a.) •|F12| or F12 = k|q1q2| / r² -F12 = [(8.99 X 10^9)(1.6 X 10^-19)(3.2 X 10^-19)]/[(0.02)²] = 1.15 X 10^-24 N •E12 = F12 / q1 -E12 is the electric field at the location of point 1, by the source charge 2, it has magnitude and direction -We can draw this on a single axis(Note: Their electric charges are radial), Charges 1 and 2 are repulsive, because both positive, so are going left and right(away from each other), E12 = -7.19 X 10^-6 N/C -q1 is the test charge, q2 is the source charge, You divide by the test charge to get the charge of the field generated by the source charge -The Electric Field(E) = (kq2/r²), or F/q, This is just the force without the test charge(q1), so it depends only on the source charge -This field is there whether or not the test charge is there, It is radial, a 360° field, If closer to the test charge, bigger force vectors pushing it away(we use a positive charge defined as the source charge), this radiates all around it, disturbed when the (+)test charge is put into the field, We looked at images of different fields(in notes) For b.) •F1 = F12 + F13(along a line) -For F12, we are imagining "3" to be missing, For F13, we are imagining "2" to be missing -These are independent forces, act as if each other are not there -F12 = 1.15 X 10^-24 N from earlier -F13 = ? -F13 = [(8.99 X 10^9)(1.6 X 10^-19)(-3.2 X 10^-19)]/[(3/4)(2 X 10^-2)]² = 1.15 X 10^-24 N -This is just: (1)/(3/4)² X F12 -F13 = (4/3)² X F12 -F13 = (16/9) X F12 = 1.78 X F12 F13 = (1.78)(1.15 X 10^-24 N) = 2.04 X 10^-24 N to the right •F1 = F12 + F13 -F1 = -1(F12) + 1.78(F12), F12 is to the left(++ = repulsion or negative direction) and F13 is to the right(+- = attraction or positive direction) -F1 = 0.78(F12) = 0.89 X 10^-24 N to the right -F1 = 8.9 X 10^-25 N to the right For c.) Like as in b.) we will use the concept of force independence, compute them independently, assuming the other is non-existent, then adding the forces together for the total net force •We predict that F12 will be repulsive and be left on the x-axis, We predict F13 will be attractive and up and to the right 60° above the x-axis, Fnet would then be the middle of these two(parallelogram method) -F1 = F12(on x-axis) + F13(at θ = 60° to x-axis), in a plane, We draw a coordinate system, +x to the right and +y up -F12 = -1.15 X 10^-24 N("-" means to the left, this is from part a.) -F13 = 1.78(F12) = 2.04 X 10^-24 N(this is to the right, this is from part b., it has the same magnitude, just a different direction from part b.) •Adding Force Components(x,y): -F12: -F12, 0 -F13: F13cosθ, F13sinθ -F1: ?, ? -To calculate F1 we will use the Pythagorean theorem to get the magnitude, then use inverse tangent to get the angle with respect to the x-axis •F1x = -F12 + F13cosθ -F1x = -F12 + 1.78(F12)cos60° -F1x = -0.11(F12) -F1x = -0.126 X 10^-24 N •F1y = 0 + F13sinθ -F1y = 0 + 1.78(F12)sin60° -F1y = 1.78(0.866)F12 -F1y = 1.77 X 10^-24 N •Pythagorean Theorem for Magnitude -F1² = F1x² + F1y² -F1 = √(F1x² + F1y²) -F1 = √[(-0.126 X 10^-24)² + (1.77 X 10^-24)²] = 1.784 X 10^-24 N •Finally, the angle -Tanθ = |F1y|/|F1x| -θ = tan-1(F1y/F1x) -θ = tan-1(1.77 X 10^-24/0.126 X 10^-24) = 86° -This is a pretty steep angle -tanθ = O/A or y/x and θ = tan-1(y/x), where tan = sin/cos

HW #1 Four identical metal spheres have charges qA = -8.0 μC, qB = -2.0 μC, qC = +5.0 μC, qD = +12.0 μC, (a.) Two of the spheres are brought together so they touch, and then they are separated, Which spheres are they, if the final charge on each is +5.0 μC?, (b.) In a similar manner which three spheres are brought together and then separated, if the final charge on each of the three is +3.0 μC?, (c.) The final charge on each of the three separated spheres in part (b) is +3.0 μC, How many electrons would have to be added to one of these spheres to make it electrically neutral?

Four Charged Spheres •qA = -8.0 μC •qB = -2.0 μC •qC = +5.0 μC •qD = +12.0 μC a.) Two spheres brought together, then separated, +5 μC/sphere •Which sphere? -Qtot = +10 μC -ΔQ = 0 → Qi = Qf = 10 μC -qB(-2 μC) and qD(+12 μC) b.) Three...+3 μC, +3 μC, +3 μC -Qtot = +9 μC -qA(-8 μC), qC(+5 μC), and qD(+12 μC) c.) How many electrons added to make each e- neutral? -Q = Ne(qe) → Ne = Q/qe = (-3 X 10^-6)/(-1.9 X 10^-19) = 1.9 X 10^13

Electrical Forces and Fields

Similar to gravitational forces and fields •We have a positive charge(q1) and a negative charge(q2), We will label q1 as the test charge and q2 as the source charge, There is an attractive force drawing q1 to q2, r is the distance between them -Fon1by2, There is also Fon2by1, same in magnitude, opposite in direction due to Newton's Third Law, but we will only be talking about the test charge -We are taking the absolute value or magnitude of the force -Coulomb's Law: F12 = k |q1q2| / r², r is very sensitive, if a little further, the force is much smaller, Once again we only care about the magnitude of the product of q1 and q2 •These forces are vectors, Have to consider trigonometry again, force in both the x- and y-directions, There is an x- and a y-coordinate plane -Looking at our diagram, F12 = 10 N, but it is at an angle -We draw y down(F12y), and x to the left(F12x), they form an angle θ = 60°, with the resultant or magnitude connecting head to tail, We can also look at in another way in which Φ = 30° and can draw it the other method, in which the magnitude splits them in the middle, Note: Based on geometry, the opposite angle of the bisector triangle has the same angle size -F12x = (10 N)(cos60°) = 5 N -F12y = (10 N)(sin60°) = 8.66 N •They add like vectors: C = A + B -Cx = Ax + Bx -Cy = Ay + By -We will do more complicated problems involving trigonometry next

The concept of force we have discussed in the past

Will be used frequently in this section

HW #2 The drawing shows and equilateral triangle, each side of which has a length of 2.00 cm, Point charges are fixed to each corner, as shown, The 4.00 μC charge experiences a net force due to the charges qA and qB, This net force points straight downward(qA = qB) and has a magnitude of 405 N, Determine the magnitudes and algebraic signs of the charges of qA and qB, Look at diagram in notes

•Charge Triangle, Note: There are 180° in a triangle! -Look at drawing in notes -a = 2.00 cm = 2.00 X 10^-2 m -Fnet = 405 N downward -qA = qB, otherwise Fnet would not straight down(same distance, so same magnitude) -qA and qB are negative, since Fnet is down •Fnet = 2Fcos30° -If we split it into two right triangles, we can do trig to find the force downwards -F = kqAq / r², where r = a, (minus sign for attractive force) •Combine: Fnet = -2(kqAq/r²)cosθ -So...qA = [-(Fnet)(a²]/[2kqcosθ] -qa = [-(405)(2.00 X 10^-2)²]/[(2)(8.99 X 10^9)(4 X 10^-6)(cos30°)] -qA = -2.6 X 10^-6 C(1 μC / 10^-6 C) -qA = qB = -2.6 μC

HW #4 The total electric field E consists of the vector sum of two parts, One part has a magnitude of E1 = 1200 N/C and points at an angle θ1 = 35° above the +x-axis, The other part has a magnitude of E2 = 1700 N/C and points at an angle of θ2 = 55° above the +x-axis, Find the magnitude and direction of the total field, Specify the direction angle relative to the +x-axis

•Electric Field -Look at drawing in notes -|E1| = 1200 N/C, θ1 = 35° -|E2| = 1700 N/C, θ2 = 55° •Find Etot = E -Ex = E1x + E2x -Ex = E1co2θ1 + E2cosθ2 -Ex = 1200 cos35° + 1700 cos55° = 1958 -Ey = E1y + E2y -Ey = E1sinθ1 + E2sinθ2 -Ey = 1200 sin35 ° 1700 sin55° = 2081 •|E| = √(1958² + 2081²) = 2858 N/C •θ = tan-1(Ey/Ex) -θ = tan-1(2081/1958) = 47°

HW #5 The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side, The initial speed of the electron is 7.00 X 10^6 m/s, The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm, Assume that the electric field between the plates is uniform everywhere and find its magnitude, Look at diagram in notes

•Electron & Capacitor -Look at drawings in notes -v(electron) = 7.00 X 10^6 m/s -Assume uniform electric field, electric field lines equally spaced •This is analogous to projectile motion -x = x0 + vx0 t + 1/2 ax t², x0 = 0, ax = 0 -y = y0 + vy0 t + 1/2 ay t², y0 = 0, vy0 = 0 (1) x = vt (2) y = 1/2ay t² -Sample point(x0, y0) = (0.02, 0.0015)m (3) ay = Fy / me = (qeEy)/me = qeE / me •Combine (1-3) -y = 1/2(qeE/me)(x/v)² -E = (2y0)(me/qe)(v/x0)², where (x,y) → (x,y0) -E = (2 X 0.0015 m)(9.11 X 10^-32 kg / 1.60 X 10^-19 C)(7.00 X 10^6 m/s / 0.02 m)² -E = 2092 N/C

Mini Electric Show Demonstration Video, Look at image in notes

•He had pieces of sticky tape, each got a charge when he stuck two together and ripped them in half after placing on the piece of glass(insulator), the one on the left side became negatively charged and the one on the right became positively charged •Electrophorus Image used to explain this demonstration -He held an insulated handle with an aluminum plate which is originally neutral -He rubbed a rabbit's fur(lots of free electrons) on the other plate, which gave it a negative charge -He then brought the aluminum plate(conductor, charge freely flows) to the negatively charged plate, which caused its electrons to go to the top(repulsion) and the protons to the bottom(attraction), this is known as polarization(like in atoms), This is called charge by induction(do not actually contact, although very close, not touching microscopically), this is what happens in thunderstorms, electrically neutral clouds push towards a charge on the ground or another cloud, vice versa, causing a lightning strike -He then used finger to touch aluminum plate which acts as a ground and takes the electrons away making the aluminum plate positive, he got a few shocks, this is conduction(actually come in contact), this is like when putting two containers of gases together and separating between one contained -He then brought the aluminum plate to the pieces of tape, where it repelled the left-positive one and attracted the right-negative one

HW #3 A small spherical insulator of mass 8.00 X 10^-2 kg and charge +0.600 μC is hung by a thread of negligible mass, A charge of -0.900 μC is held 0.150 m away from the sphere and directly to the right of it, so the thread makes an angle θ with the vertical(see the drawing), Find (a.) the angle θ and (b.) the tension in the thread

•Hung by a Thread -Look at drawings of the system and the free body force diagram -Insulator, m1 = 8.00 X 10^-2 kg, q1 = +0.600 μC, q2 = -0.900 μC, r = 0.150 m, Find θ and T •Fnet = 0 (Dynamical Equilibrium) -ΣFx = 0 -ΣFy = 0 (1) x: Fe - Tx = 0 (2) y: Ty - W = 0 (3) Fe = (kq1q2)/r² (4) Tx = Tsinθ (5) Ty = Tcosθ •Combining (1, 2) and (4, 5) -[Tsinθ = Fe]/[Tcosθ = W] = tanθ = Fe/W •Including (3) yields: tanθ = [(kq1q2)/r²]/[w] -tanθ = (1/W)[(kq1q2)/r²] -tanθ = [1/(8.00 X 10^-2 kg)(9.8 m/s²)][(8.99 X 10^9)(0.600 X 10^-6)(0.900 X 10^-6)/(0.150)²] -tanθ = 0.275 -θ = 15.4° •Ty = Tcosθ -T = Ty/cosθ = W/cosθ = 0.784/cos(15.4°) -T = 0.813 N

Ranking Tasks #1: Two Electric Charges - Electric Force Given below are seven arrangements of two electric charges, In each figure, a point labeled P is also identified, All of the charges are the same size, +Q, but they can be either positive or negative, The charges and point P all lie on a straight line, The distances between adjacent items, either between two charges or between a charge and point P, are all x cm, There are no other charges in this region, For this problem, we are going to place a charge of equal magnitude at point P, Rank these arrangements from strongest to weakest on the basis of the strength(magnitude) of the electric force on the charge when it is placed at point P, That is, put first the arrangement that will exert the strongest force on the charge at point P, and put last the arrangement that will exert the weakest force on the charge when it is placed at point P Look at image in notes, A: -, P, - B: +, -, P C: -, P, + D: +, +, P E: -, -, P F: +, P, + G: -, +, P

•Let's pretend P has a negative charge, and simplify the force as charge/distance², + or - on the force indicates the direction, not the charge A: They would both repel with the same force, so Net force = 0 units B: For the positive charge two spaces away: -1/2² = -1/4, For the negative charge one space away: +1/1² = +1, So -1/4 + 1 = 3/4, With the magnitude being Net Force = 3/4 units to the right C: For the negative charge one space away: +1/1² = 1, For the positive charge one space away: +1/1² = 1, So 1 + 1 = 2, With the magnitude being Net force = 2 units to the right D: For the positive charge two spaces away: -1/2² = -1/4, For the positive charge one space away: -1/1² = -1, So -1/4 - 1 = -5/4, With the magnitude being Net force = 5/4 units to the left E: For the negative charge two spaces away: +1/2² = +1/4, For the negative charge one space away: +1/1² = +1, So +1/4 + 1 = +5/4, With the magnitude being Net force = 5/4 units to the right F: They will both attract with the same force, so Net force = 0 G: For the negative charge two spaces away: +1/2² = +1/4, For the positive charge one space away: -1/1² = -1, So +1/4 - 1 = -3/4, With the magnitude being Netforce = 3/4 units to the left •Ranking of forces from greatest to least: C > D = E > B = G > A = F

Electric Force

•Similar to gravitational force, it is a close analogy -We have a positive charge(proton) and a negative charge(electron), this works the same, electrostatics, they pull on each other with forces of the same magnitude but opposite directions -"+-" attract and "++" or "--" repel -Q or q = the charge, they can have different amounts of charge, For example, Lithium would have a 3+ charge and 3 electrons would attach -This is a two charge model(either + or -), There are other forces besides this and gravity, strong nuclear forces are an example of a three charge model(red, blue, green), There are plenty of different forces out there -Fattraction = kqQ / r², F is in coulombs[C], r² is in [m²] is the separation distance like with gravity, k is the constant to get the units right, coulombs, This is basically the same concept as gravitational force

Gravitational Force

•There is the earth and the moon -Force from the center of earth pulling on the moon and force from the center of the moon pulling on the earth, These are forces of attraction together, momentum of the moon, orbits around the earth -These forces are equal in magnitude but opposite in direction -What is the charge for gravity? masses, M and m, this is a single charge model, F = GmM / r² -The units without G are [kg/m²], so add G, Newton's Gravitational Constant, which gets us in the units of Newtons -F vs. r graph, As r increases(further apart), F decreases exponentially(less pulling force on each other), Likewise, if closer in distance, they have stronger gravitational pulls on each other, force is very sensitive to r

Problem 5.1: Nuclear Repulsion The nucleus of an iron(Fe) atom has a radius of about 4.0 X 10^-15 m and contains 26 protons, What is the magnitude of the repulse electrostatic force between the two protons that are separated by 4.0 X 10^-15 m?

•We drew a diagram, There is a nucleus with 26 protons, also that many neutrons, There are two that are a radius away and we are calculating the repulsive force between them -k = 8.99 X 10^9 Nm²/C² -Charge of Proton = q1 = q2 = 1.6 X 10^-19 C -r = 4 X 10^-15 m •F12 = kq1q2 / r² -F12 = [(8.99 X 10^9)(1.6 X 10^-19)(1.6 X 10^-19)]/[(4 X 10^-15)²], Don't forget the radius in the denominator is squared -F12 = [(8.99 X 10^9)(1.6 X 10^-19)(1.6 X 10^-19)]/[16 X 10^-30], When exponents are squared, we times it by 2 -Let's separate the exponents from the others -F12 = [(8.99 X 1.6 X 1.6)]/16 X [(10^9)(10^-19)(10^-19)]/10^-30 -F12 = 1.44 X 10^1 N = 14.4 N -It is about 14 N which 1 N is approximately 1/4 lb., So about the force of 3.5 lbs. from this very small distance -What holds this atom together?, It is not the force of gravity(1/1000.. N), It is the strong nuclear force, it keeps the atom from exploding apart, this is the force that is involved in nuclear explosions, etc.

Concept of a Field

•We have a positive source charge -We put a negative charge in and gets attracted to it, there are invisible signals being sent communicating to each other to attract, Even if it is not there, the invisible field still exists -We can add in a positive charge known as a test charge, it would be repelled, but shows the field is already there even without the negative charge -F = kQq / r² -kQ / r² can be lumped together and called the electric field(E), So F = E(qtest), where qtest is the test charge, this explains action at a distance, Anything can be used as a source charge, we just typically pick the biggest, which in our case is the positive charge(protons in atomic nucleus)