Two small plastic spheres are given positive electrical charges. When they are a distance of 14.1 cm apart, the repulsive force between them has a magnitude of 0.205 N .
1. What is the charge on each sphere if the two charges are equal?
2. What is the charge on the first sphere if it equals one-quarter the charge of the second sphere?
3. What is the charge on the second sphere described in the previous part?
One Charge = q1 = q
Force = F = 0.205 N
Other charge = q2 =q
Distance = r = 14.1 cm = 0.141m
permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)
1.)
Using Coulomb's law,
F=[1/4pieo]q1q2/r^2
F = [1/4pieo]q^2 / r^2
q^2 =F [4pieo]r^2
q = sq rt [ 0.205[4pi8.85×10^−12 ] (0.141*0.141)
q =0.673*10^-6C
Each charge is 6.73*10^-7C ..Answer.
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2.)
One Charge = q1 = q
Force = F = 0.205 N
Other charge = q2 =q1/4
Distance = r = 0.141 m
permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)
Using Coulomb's law,
F=[1/4pieo]q1q2/r^2
q1q2 =F [4pieo]r^2
(q)(q/4) = F[4pieo]r^2
q = sq rt { 0.205[4pi8.85×10^−12 ] (0.141*0.141)*4}
q = 1.35*10^-6 C
Therefore,charge on first sphere =q/4 =1.35*10^-6C =3.3710^-7 C
..Answer.
If the charge on first sphere is 1 fourth times the charge of the
other.
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C)The charge on the second sphere is 1.35×10^-6 C ...Answer.
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Two small plastic spheres are given positive electrical charges. When they are a distance of 14.1...
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