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Q. What is the force between two small charged spheres having charges of 2x10-7 C and 3x10-7 C placed 30 cm apart in air?
Sol.
Given:
Q1 = 2x10-7 C
Q2 = 3x10-7 C
r = 30x10-2 m
We know,
F=(Q_1 Q_2)/(4πε_o r^2 )
Substituting the given values, we get
F = 6x10-3 N
Q. The electrostatic force on a small sphere of charge 0.4 micro coulomb due to another small sphere of charge -0.8 micro coulomb in air is 0.2 N.
a) What is the distance between the two spheres?
b) What is the force on the second sphere due to the first?
Sol.
a) From the formula
F=(Q_1 Q_2)/(4πε_o r^2 )
Substituting the values, we get
r = 0.12 m = 12 cm
b) Force on second sphere due to first
F21 = (Q_1 Q_2)/(4πε_o r^2 )
This comes to be -0.2 N, negative sign implying the attractive nature of force. This conforms to Newton’s third law.
Q. Check that the ratio ke2/Gmemp is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Sol.
We know very well
k = 9x109 Nm2/C2
e = 1.6x10-19 C
G = 6.6x10-11 Nm2/kg2
me = 9.11x10-31 kg
mp = 1.6x10-27 kg
Ratio = (Nm^2)/C^2 .C^2.〖kg〗^2/〖Nm〗^2 .1/〖kg〗^2 = dimensionless
And value = 2.4x1039
The ratio signifies that electrical forces are immensely stronger than the gravitational forces.
