NCERT solution|chapter 1|electric field:
Question
1.1:
What
is the force between two small charged spheres having charges of 2 × 10 -7C and
3×
10−7 C placed 30 cm apart in air?
Answer:
Electrostatic force between the spheres is given by ,
Where,
∈0 = Permittivity of free
space,
Hence,
force between the two small charged spheres is 6 × 10-3 N
Hence, force between them will be repulsive
because charges are of same nature.
Question
1.2:
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 sphere?
(b).What
is the force on the second sphere due to the first?
Answer:
f=0.2N, q1=0.4 μC =.4 x 10-6 C ,q2=− 0.8 μC =0.8x10-6 C
Electrostatic
force between the spheres is
distance
between the two spheres is 0.12 m.
Both
the spheres attract each other with same force then force for
second
sphere due to the first F is 0.2 N(as given)
Question 1.3: Check that the ratio ke2/G me and determine the value of this ratio. What does the ratio
signify ?
Answer:
as given in question,
Where,
G = Gravitational constant, Its unit is N m2 kg−2.
,me and mp = Masses of electron and proton. Their unit is kg. e = Electric charge., Its unit is C.
∈0 = Permittivity of free space, Its unit is N m2 C−2
Hence it 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 is
Question
1.4:
Fine work keep it up good work 😅😅
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