Answer:
a) F = 0.0561 N
b) F = 2.86*W
Explanation:
a) The magnitude of the electric force between the plastic spheres is given by the following formula:
[tex]F=k\frac{q_1q_2}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = q2: charge of the plastic spheres = -50.0nC = -50.0*10^-9 C
r: distance between the plastic spheres = 2.0 cm = 0.02 m
You replace the values of the parameters in the equation (1):
[tex]F=(8.98*10^9Nm^2/C^2)\frac{(-50.0*10^{-9}C)^2}{(0.02m)^2}\\\\F=0.0561N[/tex]
The electric force between the spheres is 0.0561 N
b) To calculate the relation between weight and electric force, you first calculate the weight of one of the spheres:
[tex]W=mg[/tex]
m: mass = 2.0g = 2.0*10^-3 kg
g: gravitational acceleration = 9.8 m/s^2
[tex]W=(2.0*10^{-3}kg)(9.8m/s^2)=0.0196N[/tex]
The ratio between W and F is:
[tex]\frac{F}{W}=\frac{0.0561N}{0.0196N}=2.86\\\\F=2.86W[/tex]
The electric force is 2.86 times the weight
(a) The magnitude of the electric force on each sphere is [tex]5.625 \times 10^{-2} \ N[/tex].
(b) The electric force on a sphere is larger than its weight by 2.87.
The given parameters:
The magnitude of the electric force on each sphere is calculated as follows;
[tex]F = \frac{kq^2}{r^2} \\\\F = \frac{9\times 10^9 \times (5 0 \times 10^{-9})^2}{(0.02)^2} \\\\F = 5.625 \times 10^{-2} \ N[/tex]
The weight of a sphere is calculated as follows;
[tex]W = mg\\\\W = 0.002 \times 9.8\\\\W = 0.0196 \ N[/tex]
Compare the electric force and the weight of a sphere;
[tex]= \frac{F}{W} = \frac{5.625 \times 10^{-2}}{0.0196} = 2.87[/tex]
Thus, the electric force on a sphere is larger than its weight by 2.87.
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