linear charge density of a uniformly charged straight wire of infine lenght is 10.0× 10^-8C/m calculate electric Field intensity at a distance of 2 cm from the wire
Answers
Given:
Repulsive force of magnitude 6 × 10−3 N
Charge on the first sphere, q1 = 2 × 10−7 C
Charge on the second sphere, q2 = 3 × 10−7 C
Distance between the spheres, r = 30 cm = 0.3 m
Electrostatic force between the spheres is given by the relation:
F = (1/4πε0). (q1q2)/ (r2)
Where, ε0 = Permittivity of free space and (1/4πε0) =9 × 109 Nm2C−2
Therefore, force F = (9 × 109 × 2 × 10−7)/ ((0.3)2)
= 6 × 10−3N
Hence, force between the two small charged spheres is 6 × 10−3 N. The charges are of same nature. Hence, force between them will be repulsive.
Question2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC 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?
Answer:
(a) Electrostatic force on the first sphere, F = 0.2 N
Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C
Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C
Electrostatic force between the spheres is given by the relation:
F = (1/4πε0). (q1q2)/ (r2)
Where, ε0 = Permittivity of free space and (1/4πε0) = 9 × 109 Nm2C−2
Therefore, r2 = ((1/4πε0)). ((q1q2)/ (F))
= (0.4 × 10−6 × 8 × 10−6 × 9 × 109)/ (0.2)
= 144 × 10−4
⇒ r = √ (144 × 10−4)
= 12 × 10−2
= 0.12 m
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
Question 3.
Check that the ratio ke2 /G me mp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer:
The given ratio is (ke 2)/ (G me mp)
Where, G = Gravitational constant. Its unit is N m2 kg−2
me and mp = Masses of electron and proton and their unit is kg.
e = Electric charge. Its unit is C.
k = (1/4πε0) and its unit is N m2 C−2
Therefore, unit of the given ratio
(ke 2)/ (G me mp) = ([Nm2C−2] [C−2])/ ([Nm2kg−2] [kg] [kg])
= M0L0T0
Hence, the given ratio is dimensionless.
e = 1.6 × 10−19 C
G = 6.67 × 10−11 N m2 kg-2
me= 9.1 × 10−31 kg
mp = 1.66 × 10−27 kg
Hence, the numerical value of the given ratio is:
(ke 2)/ (G me mp)
= (9 × 109 × (1.6 × 10−19)2)/ (6.67 × 10−11 × 9.1 × 10−31 × 1.67 × 10−27)
≈ 2.3 × 1039
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant
Answer:
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