Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm what is the mutual force of electrostatic
repulsion if the charge on each is is 6.5 into 10 raise to power minus 7 coulomb the radius of the radius of A and B are negligible compared to the distance of separation what is the force of repulsion if each sphere is charged double the above amount and the distance between them is half? (NCERT)
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(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10
−7
C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
(a)
Charge on sphere A, q
A
=6.5×10
−7
C
Charge on sphere B, q
B
=6.5×10
−7
C
Distance between the spheres, r=50cm=0.5m
Force of repulsion between the two spheres,
F=
4π∈
0
r
2
q
A
q
B
Where, ∈
0
= Free space permittivity
4π∈
0
1
=9×10
9
Nm
2
C
2
∴F=
(0.5)
2
9×10
9
×(6.5×10
−7
)
2
=1.52×10
−2
N
Therefore, the force between the two spheres is 1.52×10
−2
N
(b)
After doubling the charge,
Charge on sphere A, q
A
=2×6.5×10
−7
C=1.3×10
−6
C
Charge on sphere B, q
B
=2×6.5×10
−7
C=1.3×10
−6
C
Now, if the distance between the sphere is halved, then
r=
2
0.5
=0.25m
Force of repulsion between the two sphere,
F=
4π∈
0
r
2
q
A
q
B
=
(0.25)
2
9×10
9
×1.3×10
−6
×1.3×10
−6
=0.24336N
Therefore, the force between the two sphere is approximately 0.243 N.
Explanation:
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