Solved Problems-Ch23-Electric Forces New Fall 2015-2014
Solved Problems-Ch23-Electric Forces New Fall 2015-2014
Solved Problems-Ch23-Electric Forces New Fall 2015-2014
Solution
The mass of one electron is 9.11 1031 kg, so that a mass M = 75.0 kg contains
N=
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.]
Solution
(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:
(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):
And the number of electrons in the penny was (originally) 29 times this number,
(
So the fraction of electrons removed in giving the penny the given electric charge is
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3. A comb with -2.0 C of charge is 0.15m to the left from a hair with 3.0 C of charge.
Determine the force the hair exerts on the comb.
Solution
)(
4.
Solution
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5. (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?
Solution
(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?
Solution
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(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?
Solution
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6. 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?
Solution
Take for example the lower left charge in the above triangle:
7. 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.
Solution
We can use either one of the equilibrium conditions:
Lets take for example the x- component of the force on the lower left charge:
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8. 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
( )
( )
5 mm
(
Magnitudes: | |
)(
(
= 0.69
N
(
| |
)(
(
)
)
Projections:
=1.29N 1.3N
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9. Two 2.0 g charged balls hang from light weight 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
Thus,
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10.
The figure below shows two particles fixed in place: a particle of charge q1= +8q at the origin and a particle o
charge q2= 2q at x = L. At what point (other than infinitely far away) can a proton be
placed so that it is in equilibrium (the net force on it is zero)?
Page 8
Solution
Lets call Q =+3.00 nC and q = 2.00 nC and r = 4.00 cm = 0.040 m
Then,
(
)
)
(
to the right
)(
)
(
2. A uniformly charged insulating rod of length 14.0 cm is bent into the shape of a semicircle as shown in
the figure . The rod has a total charge of +7.50 C. Find
(a) the magnitude and
(b) the direction of the electric field at 0, the center of the semicircle.
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In order to find the electric field at a certain location we must add (vectorially) electric fields created by
all "point charges" in the body.
A contribution dE to the electric field, due to a differential fragment dl, can be found from Coulomb's
law.
Where
(
( )
dq
(8-x)
P
)
]
(
)(
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( )
5m
( )
3m
( )
( )
+
(
)(
5. A 16-nC charge is distributed uniformly along the x axis from x = 0 to x = 4 m. Which of the
following integrals is correct for the magnitude (in N/C) of the electric field at x = +10 m on the
x axis? Justify your answer.
4
0
36dx
(10 x )2
a.
e. none of these
b.
4
0
154dx
(10 x )2
c.
x
(
4
0
36dx
x2
d.
10
6
154dx
x2
(10-x)
x
dq
)(
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6. A particle (mass = 5 g, charge = 40 mC) moves in a region of space where the electric field is uniform and
is given by Ex = 2.3 N/C, Ey= Ez = 0. If the position and velocity of the particle at t = 0 are given by x = y =
z = 0 and vz = 20 m/s, vx = vy = 0, what is the distance from the origin to the particle at t = 2 s?
Answer. 54 m
To find the distance from the origin at t =2s, we need to find the magnitude of:
| |
(
) (
)
(
(
)(
) m
7. Two small spheres, each of mass 0.002 kg, are suspended by light strings 0.1 m in length. A uniform
electric field is applied in the x direction, as shown in the figure. The spheres have charges equal to 5.00
108 C and + 5.00 108 C. Determine the electric field that enables the spheres to be in equilibrium at an
angle = 10. Ans: 442kN/C
Lets draw the free body diagram for any of the charges and let it be the positive charge.
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( )
)
( )
(
(
)
)(
8. A proton is projected in the positive x-direction into a region of a uniform electric field
E= (- 6.00 x105) i N/C at t= 0.
The proton travels 7.00 cm as it comes to rest. Determine
(a) the acceleration of the proton,
(b) its initial speed, and
(c) the time interval over which the proton comes to rest.
Answer. 5.75x1013 m/s2, 2.84x106i m/s , 4.93x10-8 s
Solution
(a)
)
)
( )
(
(
)(
)(
)
)
( )
Page 13
9. Negative charge is distributed uniformly around a quarter-circle of radius that lies in the
first quadrant, with the center of curvature at the origin. Find the components of the net
electric field at the origin.
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10. A Positive charge is distributed none uniformly around a quarter-circle with =osin3 and of radius R
that lies in the first quadrant, with the center of curvature at the origin. Find the net electric field at the origin
=osin3
(
( )
Note:
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11. The figure (a) shows a plastic rod having a uniformly distributed charge Q. The rod has been
bent in a 120 circular arc of radius r. We place coordinate axes such that the axis of symmetry
of the rod lies along the x axis and the origin is at the center of curvature P of the rod. In terms of
Q and r, what is the electric field due to the rod at point P?
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12. Semi circle of radius a is in the first and second quadrants, with the center of curvature at the origin.
Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge is
distributed uniformly around the right half of the semicircle. What are the magnitude and direction of the
net electric field at the origin produced by this distribution of charge?
Solution: We must add the electric field components of the positive half and the negative half. From
Problem 8, the electric field due to the quarter-circle section of positive charge has components
The field due to the quarter-circle section of negative charge has components
The resultant field is the sum of the x- and y-components of the fields due to each half of the semicircle.
The y-components cancel, but the x-components add, giving
in the +x-direction
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13. A thin plastic rod is bent so that it makes three quarters of a circle with radius R as shown in the figure.
If two quarters of the circle carry a uniform charge density + and one quarter of the circle carries a
uniform charge density -, what is the magnitude of the electric field at the center?
Solution:
First calculate the electric field at the center of a semi-circle with charge density +
Now calculate the electric field at the center of a quarter-circle with charge density as follows:
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14. Four parallel infinite sheets of charge spaced by 4 cm between adjacent sheets are shown edgewise
in the figure. Their surface charge densities are as indicated with = 2.0C/m2. What is the magnitude
of the electric field (in N/C) at point P located midway between two sheets as shown? (Hint: Electric
fields are vectors that add as such.) Answer: 4.5 105N/C
Solution:
Sheet (A) produces an electric field of magnitude /(2o) going away from it, which means the
electric field is pointing towards the bottom of the page at point P. Sheet (B) produces an
electric field of magnitude 2/(2o) going towards it, which implies that the electric field is
pointing towards the bottom of the page at point P. Similarly, at point P the electric fields due
to sheets C and D are 3/(2o) toward the top of the page and 4/(2o) toward the bottom
of the page. Adding these four fields together produces a net electric field of 4/(2o) toward
the bottom of the page.
15. Two thin 80.0 cm rods are oriented at right angles to each other. Each rod has one end at the origin of the
coordinates, and one of them extends along the +x-axis while the other extends along the +y-axis. The rod
along the +x-axis carries a charge of 15.0 C distributed uniformly along its length, and the other rod
carries +15.0 C uniformly over its length. Find the magnitude and direction of the net electrical force
-19
that these two rods exert on an electron located at the point (40.0 cm, 40.0 cm). (e = 1.60 10 C, O = 8.85
-12
10
C /N m ) Answer: 1.35 10
-13
Solution:
The net electric field at point (p) is the resultant of both fields for each finite charged wire
(
)(
)
Page 19
Also
(
)(
| |. Find
16. A finite none conducting line charge of length (2b) with varying linear charge density
the electric field at a at point P on the perpendicular bisector of the rod at distance (a) as shown in the figure.
| |
As, can be seen from the figure above, that the x- components will cancel each other due
to the symmetry of the line charge around the origin. Therefore:
[
Thus;
)
(
(
)
| |
(
) (
Page 20
17. A spherical water drop 1.20 m in diameter is suspended in calm air owing to a downwarddirected
atmospheric electric field E = 462 N/C
(a) What is the weight of the drop?
(b) How many excess electrons does it have?
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