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Questions: 1 2 3 4 5 6 7 8 9 10 11

Coulomb's Law and Electric Fields Solutions

  1. Copper has a density of 8.890 × 103 kg/m3, an atomic number of 29, and an atomic mass of 63.546 g. Determine the number of electrons per cubic centimetre in a chunk of copper. Atomic number is the number of protons, and hence the number of electrons, in a neutral atom. Atomic mass is the mass of one mole (6.0221 × 1023) of the particular atoms in question.

    2. We need to know how many moles of copper, NCu, is contained in 1 cm3. Then the number of electrons is Ne = Z Ncu. To find the number of moles we fist find the mass of 1 cm3 in grams.

    m = 1 cm3 × 8.890 × 103 kg/m3 × (1000 g/kg) × (1 m / 100cm)3 = 8.890 g .

    The number of moles is this mass divided by the molar mass

    NCu = m/Matomic = 8.890 g / 63.546 g/mol = 0.1399 mol .

    Hence the number of electrons is

    Ne = Z NCu = 29 × 0.1399 mol × 6.0221 × 1023 /mol = 2.44 × 1024 .

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  2. There are three identical metal spheres, A, B, and C. Sphere A carries a charge of +5q. Sphere B carries a charge of -q. Sphere C carries no charge. Spheres A and B are touched together then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. How much charge ends up on sphere C?

    3. It is reasonable to assume that the total charge on two touching balls divides equally among the two balls.

    Thus a charge of 1.5q ends up on sphere C.

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  3. Where would you put a positive charge of +1 μC in the diagram below so that the net electrostatic force on it is zero?

    4. We would have to put the 1 μC where the force from the right 2.0 μC charge is cancelled by the force from the left 5.0 μC charge. Since forces are vectors, we would have to put the charge somewhere on the line joining the two charges as shown below. We will assume that the 1 μC charge is some distance x from the 2.0 μC charge.

    We are asked to calculate the force on an electric charge due to other electric charges. To do this we follow the following steps:

    The 1 μC charge feels a force from the 5.0 μC charge whose magnitude is . The charges have the same sign so they repel. The 1 μC charge feels force F51 directed to the right. The 1 μC charge feels a force from the 2.0 μC charge whose magnitude is . The charges have the same sign so they repel. The 1 μC charge feels force F21 directed to the left.

    We sketch the forces as shown in the diagram below

    For the net force to be zero, F21 must have the same magnitude as F51, F21 = F51. Thus we have

    .

    We eliminate the common factor kQ1 and we get

    .

    Next we take the square root of each side,

    .

    We cross-multiply, collect terms, and find

    .

    We would have to place the 1 μC charge 1.16 m from the 2 μC charge for it to feel no net force.

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  4. What is the net force on charge A in each configuration shown below? The distances are r1 = 12.0 cm and r2 = 20.0 cm.

    5. We are asked to calculate the force on an electric charge due to other electric charges. To do this we follow the following steps:

    Charge A feels a force due to charge B, the vector from B to A is r1 = (0.120 m) i. The Coulomb force FBA is thus

    Charge A feels a force FBA directed to the right in the positive i direction, that is they repel because A and B have the same sign.

    Charge A experiences a force from charge C, the vector from C to A is r2 = (0.200 m) j. The Coulomb force FCA is thus

    The charges have the opposite sign so they attract. Charge A feels force FAC directed straight down, the −j direction.

    We sketch the forces

    The net force is

    Fnet = FBA + FCA = (9.36458 N)i − (3.37125 N)j

    We use the Pythagorean Theorem to find Fnet,

    Fnet = [(FBA)2+(FCA)2]½ = [(9.36458)2+(3.37125)2]½ = 9.95 N .

    We use trigonometry to find the angle θ ,

    θ = arctan(FCA/FBA) = arctan(–3.37125/9.36458) = –19.8° .

    The net force acting on charge A due to the other forces is 9.95 N at 19.8° below the horizontal.

    The magnitudes of FBA and FCA are the same as in part (a), however, the direction of the force of charge C on charge A is different. That force is now directed to the left (–j) as shown in the diagram below.

    Since the forces are along the same axis we find

    Fnet = FBA −FCA = (9.36458 N)i – (3.37125 N)j = (5.99 N)i .

    The net force acting on charge A due to the other forces is 5.99 N along the positive x axis.

    The magnitudes of FBA and FCA are the same as in part (a), however, the direction of the force of charge C on charge A is different. That force is now directed to the right (+i) as shown in the diagram below.

    Since the forces are along the same axis we find

    Fnet = FBA + FCA = (9.36458 N) i + (3.37125 N) i = (12.74 N)i .

    The net force acting on charge A due to the other forces is 12.7 N along the positive x axis

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  5. A charged ball of mass m = 0.265 kg and unknown charge q is hanging by a light thread from a ceiling. A fixed charge Q = +5.00 μC on an insulated stand is brought close to the unknown charge. As a result, the unknown charge hangs at an angle q = 38.0° to the vertical as shown in the diagram below. The distance between the two charges is r = 22.0 cm.

    6. (a) What is the sign of the unknown charge? Explain how you know this.
    (b) What is the magnitude of the unknown charge?

    (a) The unknown charge q must be negative since it is attracted to the positive charge Q.

    (b) Notice that the unknown charge is not moving, so all the forces acting on it must balance. In dealing with forces, we draw a free body diagram and apply Newton's Second Law. Since it is not moving, the acceleration is zero. The forces acting on the unknown charge are its weight, mg, the tension in the string T, and the Coulomb force F. The magnitude of F is given by F = |kqQ/r2|.

    i
    j
    Fx = max Fy = may
    k|qQ|/r2 - Tsin(θ) = 0 Tcos(θ) - mg = 0

    The second column gives us an expression for T, T = mg/cos(θ). If we substitute this into the first equation, we find an equation, k|qQ|/r2 = mgsin(θ)/cos(θ), or

    k|qQ|/r2 = mgtan(θ).

    Solving for q, we get |q| = r2mgtan(θ)/k|Q|. Using the given values we find

    |q| = (0.22)2(0.265)(9.81)tan(38°) / (8.99×10-9)(5×10-6) = 2.19 × 10-6 C .

    So the unknown charge is q = -2.19 C.

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  6. An object has a mass of 215 kg and is located at the surface of the earth (mass = 5.98 × 1024 kg, radius = 6.38 × 106 m). Suppose this object and the earth each have identical charge q. Assuming that the earth's charge is located at the centre of the earth, determine q such that the electrostatic force exactly cancels the gravitational force.

    7.  

     
     

    Using Newton's Second Law, we have

    FCoulomb = Fgravity
    kq2/R2 = GMEm/R2
    kq2 = GMEm

    Hence we find

    q = [GMEm/k]½
    = [(6.672 × 10-11 N-m2/kg2)(5.98 × 1024 kg)(215 kg)/(8.99 × 109 N-m2/C2)]
    = 3.09 × 103 C

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  7. Three point charges q1 = −1.00 μC, q2 = 2.00 μC, and q3 = 3.00 μC are placed at the corners of an equilateral triangle of side length L = 0.250 m. Find the magnitude and the direction of the electric field at (a) a point midway between charges q1 and q2, and at (b) the centre of the equilateral triangle.

    8. We are asked to calculate the net electric field at a point due to electric charges. To do this we follow the following steps:

    Doing a little trigonometry, we find

    r1 = –[½L sin(30º)]i – [½L cos(30º)]j = –(L/4)i – (Ö3L/4)j,

    r2 = [½L cos(60º)]i + [½L sin(60º)]j = (L/4)i + (Ö3L/4)j, and

    r3 = [½L cos(60º) – L]i + [½L sin(60º)]j = –L(3/4)i + (Ö3L/4)j,

    The magnitudes of these vectors are found using the Pythagorean Theorem,

    r1 = r2 = [(L/4)2 + (Ö3L/4)2]½ = ½L,

    which is the expected result and confirms our previous work. As well

    r3 = [(3L/4)2 + (Ö3L/4)2]½ = (3/4)½L .

    Having found the directions of the electric fields, as shown in the diagram, we now determine the electric field due to each charge

    E1
    		 = kq1r1/(r1)3
      = (8.99 × 109 N-m2/C2)(–1.00 × 10-6 C)[–(L/4)i – (Ö3L/4)j ]/( ½L)3
      = (2.87680 × 105 N/C)i + (4.98276 × 105 N/C)j
       
    E2
    		 = kq2r2/(r2)3
      = (8.99 × 109 N-m2/C2)(2.00 × 10-6 C)[(L/4)i + (Ö3L/4)j ]/( ½L)3
      = (5.7536 × 105 N/C)i + (9.96555 × 105 N/C)j
       
    E3
    		 = kq3r3/(r3)3
      = (8.99 × 109 N-m2/C2)(3.00 × 10-6 C)[–L(3/4)i + (Ö3L/4)j]/((3/4)½L)3
      = –(4.982764 × 105 N/C)i + (14.94829 × 105 N/C)j
    The net electric field is thus

    Enet = E1 + E2 + E3 = (3.64763 × 105 N/C)i + (17.8251 × 105 N/C)j .

    Using the Pythagorean Theorem, the magnitude of the net electric field is

    Enet = [(Ex)2 + (Ey)2]½ = 1.82 × 106 N/C .

    Using trigonometry to find the direction, we get

    θ = arctan(Ey/Ex) = arctan(17.8251/3.64763) = 78.4° .

    Thus E = (1.82 × 106 N/C, 78.4° ).

    Doing a little trigonometry, we find

    r1 = − [½L / cos(30° )]j = −L/Ö3 j,

    r2 = [½L ]i + [½L tan(30º)]j = (L/2)i + (L / 2Ö3)j, and

    r3 = − [½L]i + [½L sin(60º)]j = –(L/2)i + (L / 2Ö3)j,

    The magnitudes of these vectors are found using the Pythagorean Theorem,

    r2 = r3 = [(L/2)2 + (L/2Ö3)2]½ = L/Ö3,

    which is also the magnitude of r1 as expected from the drawing.

    Having found the vector directions, we now determine the electric field due to each charge

    E1
    		 = kq1r1/(r1)3
      = (8.99 × 109 N-m2/C2)(–1.00 × 10-6 C)[ – (L/Ö3)j ]/(L/Ö3)3
      = + (4.31520 × 105 N/C)j
       
    E2
    		 = kq2r2/(r2)3
      = (8.99 × 109 N-m2/C2)(2.00 × 10-6 C)[(L/2)i + (L/2Ö3)j ]/(L/Ö3)3
      = (7.47415 × 105 N/C)i + (4.31520 × 105 N/C)j
       
    E3
    		 = kq3r3/(r3)3
      = (8.99 × 109 N-m2/C2)(3.00 × 10-6 C)[–(L/2)i + (L/2Ö3)j]/(L/Ö3)3
      = –(11.21122 × 105 N/C)i + (6.47280 × 105 N/C)j
    The net electric field is thus

    Enet = E1 + E2 + E3 = −(3.73707 × 105 N/C)i + (15.1032 × 105 N/C)j .

    Using the Pythagorean Theorem, the magnitude of the net electric field is

    Enet = [(Ex)2 + (Ey)2]½ = 1.56 × 106 N/C .

    Using trigonometry to find the direction, we get

    θ = arctan(Ey/Ex) = arctan(15.1032/− 3.7371) = −76.1° ,

    note that we are in quadrant II so the angle is actually 180º - q . Remember to always check which quadrant the vector is in!

    Thus E = (1.56 × 106 N/C, 103.9° ).

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  8. The distances on the graph below are in metres.  The charges are Q1 = –3 μC, Q2 = 2 μC, and Q3 = 5 μC. Find the net electric field at the origin.

    We are asked to calculate the net electric field at a point due to electric charges. To do this we follow the following steps:

    The vectors distances can be easily read from the graph:

    r1 = 9i – 5j            r = (92 + 52)½ = 106½

    r2 = 7i + 4j            r = (72 + 42)½ = 65½

     r3 = –3i – 4j         r = (32 + 42)½ = 5

    The electric field due to each charge is thus:

    E1 = kQ1r1 / r13 =(8.99 × 109)(–3 × 10-6)(9i – 5j) / 1063/2 = –222.42i + 123.56j N/C

    E2 = kQ2r2 / r23 =(8.99 × 109)(2 × 10-6)(7i + 4j) / 653/2 = 240.17i + 137.24j N/C

    E3 = kQ3r3 / r33 =(8.99 × 109)(5 × 10-6)(–3i – 4j) / 53 = –1078.8i – 1438.4j N/C

    The sum of the electric fields is thus Enet = –1061i – 1178j N/C.

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  9. Find the net electric field at point A in the diagram below.

    10. We are asked to calculate the net electric field at a point due to electric charges. To do this we follow the following steps:


    Using some trigonometry, the vector distances are

    r1 = (3.0 m)i + (1.6 m)j ,

    r2 = (3.0 m)i + (2.8 m)j ,

    and

    r3 = (2.8 m)j .

    The magnitudes of these vectors are found using the Pythagorean Theorem,

    r1 = [(3)2 + (1.6)2]½ = 3.40 m,

    r2 = [(3)2 + (2.8)2]½ = 4.10366 m,

    and r3 = 2.8 m.

    Having found the vector directions, we now determine the electric field due to each charge

    E1
    		 = kq1r1/(r1)3
      = (8.99 × 109 N-m2/C2)(5.00 × 10-6 C)[(3.0 m)i + (1.6 m)j ]/(3.4 m)3
      = (3430.95 N/C)i + (1829.84 N/C)j
       
    E2
    		 = kq2r2/(r2)3
      = (8.99 × 109 N-m2/C2)(5.00 × 10-6 C)[(3.0 m)i + (2.8 m)j ]/(4.10366 m)3
      = (1951.36 N/C)i + (1821.27 N/C)j
       
    E3
    		 = kq3r3/(r3)3
      = (8.99 × 109 N-m2/C2)(2.00 × 10-6 C)[(2.8 m)j]/(2.8 m)3
      = (2293.37)j
    The net electric field is thus

    Enet = E1 + E2 + E3 = (5382.31 N/C)i + (5944.48 N/C)j .

    Using the Pythagorean Theorem, the magnitude of the net electric field is

    Enet = [(Ex)2 + (Ey)2]½ = 8019 N/C .

    Using trigonometry to find the direction, we get

    θ = arctan(Ey/Ex) = arctan(5944.48/5382.31) = 47.84° .

    Thus E = (8019 N/C, 47.84° ).

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  10. A small ball of mass m = 0.015 kg is suspended floating in an electric field of magnitude E = 5000 N/C. (a) If the electric field is pointing straight up into the air, what is the charge on the ball? (b) If E points straight down?

    11. (a) A charge in an electric field experiences a Coulomb force F = |q|E. The ball has mass, so weight acts on it. The ball is not moving so it has no acceleration. This means the weight must be balanced by the Coulomb force. Since the Coulomb force and the electric field are in the same direction, this indicates that the charge is positive.

    We apply Newton's Second Law and find |q|E - mg = 0. Upon rearranging we find

    |q| = mg/E = (0.015)(9.81)/(5000) = 2.94 × 10-5 C.

    The unknown charge is +29.4 C.

    (b) Here the electric field is straight down. However the Coulomb force must still point up to balance the weight. Since the Coulomb force and the electric field are in opposite directions, this indicates that the charge is negative.

    We apply Newton's Second Law and find |q|E - mg = 0. Upon rearranging we find

    |q| = mg/E = (0.015)(9.81)/(5000) = 2.94 × 10-5 C.

    The unknown charge is -29.4 C.

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  11. A ball of mass m = 0.010 kg and charge q is tied by a very light string to the ceiling. The effects of a uniform electric field E = 5000 N/C has caused the charged ball to move to one side so that the string makes an angle of θ = 37° with the vertical. Draw the free body diagram of the floating ball. Determine the magnitude of the charge q. How can you tell the sign (positive or negative) of the charge?

    12. The charge is hanging at an angle because it feels a Coulomb force from being in an electric field, F = |q|E. The Coulomb force is clearly to the right because that is the way the charge moved. Since the Coulomb force and E are in opposite directions, the charge must be negative.

    Since we have the Coulomb force, the weight, the tension, and know that the acceleration is zero since the body is not moving, we apply Newton's Second Law.

    i
    j
    Fx = max Fy = may
    |q|E - Tsin(θ) = 0  Tcos(θ) - mg = 0 

    From the second column, we have T = mg/cos(θ). We can substitute this in to the equation in the first column to get, |q|E - [mg/cos(θ)]sin(θ) = 0 . Solving for |q| yields

    |q| = mgtan(θ)/E = (0.010 kg)(9.81 m/s2)tan(37°)/(5000 N/C) = 1.4785 × 10-5 C .

    The unknown charge is thus, q = -14.8 C.

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