Solved Physics Homework Examples, Problems on Coulomb's Law.
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Exercise One, Homework Help Solutions. Coulomb's Law Applied to a Case of Electrical Interaction Between Equal Sign Charges.
Two small similar spheres with the same negative charge and 40 g of mass each are hanging at rest as indicated in the figure. The length of each string is 20 cm and the angle ß is 6º. Calculate how many electrons are needed to add to each sphere to obtain this situation.
The spheres distance is 2a and a is obtained from sin6º = a/0.2m, this is a = 0.021 m. The distance r between the spheres is then 0.042 m.
On the left sphere's free body diagram, where T is the string tension, Fs is the repulsed force the right sphere exerts over the left one and mg is the sphere weight. Then we have:
Sum of Fx = Tsinß - Fs = 0
2) Sum of Fy = Tcosß - mg = 0 or T = mg/cosß, which replaced in 1) we get: Fs = (mg/cosß)sinß = mgtgß = (40x10-3 kg)(9.8 m/s2)tan6º = 4.12x10-2 N.
The charge q on each sphere is calculated using Coulomb's law,
Fs = ke q2 /r2.
Replacing r for 0.042 m and Fs for 4.12x10-2 N and solving for q2, q2 = Fs r2/ke = (4.12x10-2 N)( 0.042 m)2/9x109 Nm2/C2. Then q = 8.98x10-8 C.
The number of electrons needed to add to each sphere is 8.98x10-8 C/1.602x10-19C = 5.6x1011 electrons.
Exercise Two, Homework Help Solutions. Coulomb's Law Applied to a Case of Three Point Charges
Assume three point charges located at the vertex of a right triangle as shown in the figure, where q1 = -80µC, q2 = 50µC and q3 = 70µC, distance AC = 30 cm, distance AB = 40 cm. Calculate the force over the charge q3 due to the charges q1 and q2.
The force directions coincide with the lines
joining each pair of point charges. The force that q1 exerts over
q3, F31 is of attraction. The force that q2
exerts over q3, F32 is repulsive. This is shown in the
figure where the forces F31 and F32 are vectorially
added to obtain F3. The distance between q3 and q1
is obtained from
(CB)2 = (AC)2 + (AB)2 = (0.3 m)2 + (0.4 m)2
from where CB = 0.5 m
The magnitudes of such forces are:
F31 = [(9x109 Nm2
/C2)(80x10-6 C)(70x10-6 C)]/(0.5 m)2
= 201.6 N
F32 = [(9x109 Nm2/C2)(50x10-6 C)(70x10-6 C)]/ (0.3 m)2= 350 N
It is convenient to use coordinates axis as indicated, placing the origin on the charge where we need to calculate the resultant force, in this case on q3.
The resultant force is F3 = F31
+ F32 . Then in terms of components x and y:
F3x = F31x + F32x.
F3y = F31y + F32y.
F31x = F31cosß = (201.6
N)(0.4/0.5) = 161.3 N ; F31y = - F31senß =-(201.6 N)(0.3/0.5)
= -121 N
F32 = 0 ; F32y = F32 = 350 N.
F3x= 161.3 N + 0 = 161.3 N ; F3y = -121 N + 350 N = 229 N.
The magnitude of the resultant force F3 is obtained from (F3)2 = (F3x)2 + (F3y)2, giving F3 = 280 N. The angle this force makes with the x axis is calculated using tgµ = F3y/ F3x= 229/161.3 = 1.42, then µ = arctan1.42 = 54.8º.
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