Solved Problems-Ch23A-Electric Forces New Spring 2013-2014

  • Published on
    05-Mar-2016

  • View
    213

  • Download
    0

DESCRIPTION

solved problems on electric forces

Transcript

  • Phy201-Ch23- Electric Charge and Electric Field Page 1

    Ch-23 Electric Charge and Electric Field -Problems and Solutions Phy201- Dr.Hikmat A. Hamad 1. What is the total charge of 75.0 kg of electrons?

    The mass of one electron is 9.11 1031 kg, so that a mass M = 75.0 kg contains

    N =

    = 75.09.111031 = 8.23 1031 electrons The charge of one electron is e = 1.60 1019 C, so that the total charge of N electrons is: Q = N(e) = (8.23 1031) (1.60 1019 C) = 1.32 1013 C

    2. (a) How many electrons would have to be removed from a penny to leave it with a charge of +1.0107 C? (b) What fraction of the electrons in the penny does this correspond? [A penny has a mass of 3.11 g; assume it is made entirely of copper.] (a) We know that as each electron is removed the penny picks up a charge of +1.60 1019 C. So to be left with the given charge we need to remove N electrons, where N is:

    =

    = 1.0 107 C1.60 1019 C = 6.21011 (b) To answer this part, we will need the total number of electrons in a neutral penny; to find this, we need to find the number of copper atoms in the penny and use the fact that each (neutral) atom contains 29 electrons. To get the moles of copper atoms in the penny, divide its mass by the atomic weight of copper (see periodic table): = 3.11 63.54 = 4.89 102 The number of copper atoms is

    = = 4.89 102 6.0221023 = 2.95 1022 And the number of electrons in the penny was (originally) 29 times this number,

    = 29 = 29(2.95 1022) = 8.551023 So the fraction of electrons removed in giving the penny the given electric charge is

    = 6.210118.551023 = 7.3 1013 And it is very small fraction.

  • Phy201-Ch23- Electric Charge and Electric Field Page 2

    3.

  • Phy201-Ch23- Electric Charge and Electric Field Page 3

    4. (a) The figure (a) shows two positively charged particles fixed in place on an x- axis. The charges are q1=1.60 x 10-19 C and q2=3.20 X 10 -19 C, and the particle separation is R = 0.0200 m. What are the magnitude and direction of the electrostatic force F12 on particle 1 from particle 2?

    (b) Figure (c) is identical to Fig.(a) except that particle 3 now lies on the x axis between particles 1 and 2.Particle 3 has charge q3 = -3.20 x 10-19 C and is at a distance 3/4R from particle l. What is the net electrostatic force F1,net on particle 1 due to particles 2 and 3?

  • Phy201-Ch23- Electric Charge and Electric Field Page 4

    (c) Figure (e) is identical to Fig. (a) except that particle 4 is now included. It has charge q4= -3.20 x 10-19 C, is at a distance 3/4R from particle 1, and lies on a line that makes an angle = 60o with the x axis. What is the net electrostatic force F1,net on particle 1 due to particles 2 and 4?

  • Phy201-Ch23- Electric Charge and Electric Field Page 5

    5. Three equal charges, each of +7 C are arranged at the corners of an equilateral triangle of side length 10 cm. What is the magnitude of the total force (in N) acting on each of the charges?

    Take for example the lower left charge in the above triangle:

    6. Four charges of the same sign and value q are placed in the corners of a square and free to move. One more charge Q is placed in the center of the square so that the entire system of the five charges has became statically stable, i.e. net forces on each of the five charges are equal to zero. Find the value of charge Q in terms of q.

    We can use either equilibrium conditions:

    = 0 = 0 Lets take for example the x- component of the force on the lower left charge:

  • Phy201-Ch23- Electric Charge and Electric Field Page 6

    7. If a = 3.0 mm, b = 4.0 mm, Q1 = 60 nC, Q2 = 80 nC, and q = 32 nC in the figure, what is the magnitude of the total electric force on q? Answer. 1.3 N

    Lets find the angle and r: = 1 = 1 34 = 36.9 ,

    = 2 + 2 = 5 mm Magnitudes: |1| = 12 = 9109 60109(32109)(5103)2 = 0.69 N |2| = 22 = 9109 80109(32109)(5103)2 = 0.92 Projections: 1 = 1 cos = 0.69 cos 36.9 = 0.55

    1 = 1 sin = 0.69 sin 36.9 = 0.41 2 = 2 cos = 0.92 cos 36.9 = 0.74 2 = 2 sin = 0.92 sin 36.9 = 0.55 Components of the net force:

    = 1 + 2 = 0.55 + 0.74 = 1.29 = 1 + 2 = 0.41 + 0.55 = 0.14 Magnitude of the net force:

    || = 2 + 2 = 2 + 2 =1.292 + 0.412 =1.29N 1.3N

  • Phy201-Ch23- Electric Charge and Electric Field Page 7

    8. Two 2.0 g charged balls hang from lightweight insulating threads 1 m long from a common support point as shown in the figure. If one of the balls has a charge of 0.01 C and if the balls are separated by a distance of 15 cm, what is the charge on the other ball (in C)? Answer: 0.37

    Solution: The conditions for static equilibrium on ball Q2 is

    .. (1)

    .. (2)

    And dividing the 1st equation by the 2nd yields

    Where =

    =

    Solving for the product of the two charges gives

    Thus,