there are three points A B and C drawn on the surface of a sphere they are connected to each other such the length ab BC and CA are the shortest paths drawn on the surface of the spear A particle takes a round trip such that it moves from point A to B there to see and back to a the displacement of the particle is
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The potential around a point charge is V(r)=
4πϵ
0
r
q
Let a,b,c be the radii of A,B,C respectively.
Thus, t
1
=b−a and t
2
=c−b
V
A
−V
B
=
4πϵ
0
a
q
−
4πϵ
0
b
q
=
4πϵ
0
ab
q t
1
V
B
−V
C
=
4πϵ
0
b
q
−
4πϵ
0
c
q
=
4πϵ
0
bc
q t
2
But, V
A
−V
B
=V
B
−V
C
⇒
V
B
−V
C
V
A
−V
B
=1⇒
ab
t
1
×
t
2
bc
=1⇒
t
2
t
1
=
c
a
As a<c, we have t
1
<t
2
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