Two insulated charged spheres of charges 6.5*10^-7 C each are seperated by a distance of 0.5 m . Calculate the electrostatic force between them. Also calculate the force 1) when the the charges are doubled and the distance of seperation is halved 2) when the charges are placed in a dielectric medium water (ε =80)
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PHYSICS
(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?
December 26, 2019avatar
Meghana Dungriwala
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ANSWER
(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 sphe
this is the answer
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