**Coulomb's
Law Explained, Press Here**

**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, F _{s} is the repulsed force the right
sphere exerts over the left one and mg is the sphere weight. Then we have:**

**1)
Sum of F _{x} = Tsinß - Fs = 0
**

**The charge q on each sphere is calculated
using Coulomb's law, **

**F _{s}
= k_{e} q^{2} /r^{2}.**

**Replacing r for 0.042 m and F _{s}
for 4.12x10^{-2} N and solving for q^{2}, q^{2} =
F_{s} r^{2}/k_{e} = (4.12x10^{-2} N)( 0.042
m)^{2}/9x10^{9} Nm^{2}/C^{2}. Then q = 8.98x10^{-8}
C. **

**The number of electrons needed to add to
each sphere is 8.98x10 ^{-8} C/1.602x10^{-19}C = 5.6x10^{11}
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 q _{1} = -80µC,
q_{2} = 50µC and q_{3} = 70µC, distance AC = 30
cm, distance AB = 40 cm. Calculate the force over the charge q_{3}
due to the charges q_{1} and q_{2}.**

**The force directions coincide with the lines
joining each pair of point charges. The force that q _{1} exerts over
q_{3}, F_{31} is of attraction. The force that q_{2}
exerts over q_{3}, F_{32} is repulsive. This is shown in the
figure where the forces F_{31} and F_{32} are vectorially
added to obtain F_{3}. The distance between q_{3} and q_{1}
is obtained from**

from where CB = 0.5 m

**The magnitudes of such forces are:**

**F _{31} = [(9x10^{9} Nm^{2}
/C^{2})(80x10^{-6} C)(70x10^{-6} C)]/(0.5 m)^{2}
= 201.6 N**

**F _{32} = [(9x109 Nm^{2}/C^{2})(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 q _{3}.**

**The resultant force is F _{3} = F_{31}
+ F_{32} . Then in terms of components x and y:**

**F _{31x} = F_{31}cosß = (201.6
N)(0.4/0.5) = 161.3 N ; F_{31y} = - F_{31}senß =-(201.6 N)(0.3/0.5)
= -121 N**

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