Answer:
r = 0.303m
= 30.3cm
Explanation:
Given that,
The number of electrons transferred from one sphere to the other,
n = 1 ×10 ¹³e le c t r o n s
The electrostatic potential energy between the spheres,
U = − 0.061 J
The charge on an electron,
q = − 1.6 × 10 ⁻¹⁹C
The coulomb constant,
K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²
Due to the transfer of electrons, both spheres become equally and oppositely.
The charge gained by the sphere due to the excess of the electron is:
q ₁ = n q
= 1 ×10 ¹³ * − 1.6 × 10 ⁻¹⁹
= -1.6 × 10⁻⁶C
The charge left on the first sphere is =
q ₂ = -q₁ = 1.6 × 10⁻⁶C
The electric potential energy between two point charges is given by the following equation:
U = K q ₁q ₂/r
q ₁ and q ₂ are the two charges.
r is the distance between the charge and the point.
K = 8.98755 × 10 ⁹ N ⋅ m ² / C ²
we have:
-0.061 = (8.98755 × 10 ⁹ * (-1.6 × 10⁻⁶)²) / r
r = (18.41 × 10 ⁻³) / 0.061
r = 0.303m
= 30.3cm
The distance between the two spheres is 30.3cm
The number of electrons transferred is denoted by
n = 1 ×10 ¹³e le c t r o n s
The electrostatic potential energy between the spheres,
U = − 0.061 J
The charge on an electron,
q = − 1.6 × 10 ⁻¹⁹C
The coulomb constant,
K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²
Hence, due to the transfer of electrons, both spheres become equal and opposite.
Then, the charge gained by the sphere due to the excess of the electron is:
The charge left on the first sphere is =
q ₂ = -q₁ = 1.6 × 10⁻⁶C
The electric potential energy between two point charges is given by the following equation:
U = K q ₁q ₂/r
Where
q ₁ and q ₂ are the two charges.
r is the distance between the charge and the point.
K = 8.98755 × 10 ⁹ N ⋅ m ² / C ²
Furthermore, we arrive at
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