2 spheres A and B are joined to each other by a connecting wire. A is positively charged and B is negatively charged. a. Which sphere has more electrons?
Answers
Explanation:
Let a be the radius of a sphere A, Q
A
be the charge on the sphere, and C
A
be the capacitance of the sphere. Let b be the radius of a sphere B, Q
B
be the charge on the sphere, and C
B
be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let E
A
be the electric field of sphere A and E
B
be the electric field of sphere B. Therefore, their ratio,
E
B
E
A
=
4π∈
0
×a
2
Q
A
×
Q
B
b
2
×4π∈
0
E
B
E
A
=
Q
B
Q
A
×
a
2
b
2
...(1)
However,
Q
B
Q
A
=
C
B
V
C
A
V
And
C
B
C
A
=
b
a
∴
Q
B
Q
A
=
b
a
...(2)
Putting the value of (2) in (1), we obtain
∴
E
B
E
A
=
b
a
a
2
b
2
=
a
b
Therefore, the ratio of electric fields at the surface is
a
b
.
A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.
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