Question 1.2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-46
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(a)Let charge on small sphere is denoted by q₁
Charge on large sphere is denoted by q₂
force is denoted by F
Given, q₁ = 0.4μC = 0.4 × 10⁻⁶ C
q₂ = 0 8μC = 0.8 × 10⁻⁶ C
F = 0.2 N
∴ use Coulombs law,
F = Kq₁q₂/r²
0.2 = 9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶/r²
r² = (9 × 0.32) × 10⁻³/0.2
r² = 14.4 × 10⁻³
r² = 144 × 10⁻⁴
Taking square root both sides,
r = 12 × 10⁻² m = 12cm
(b) according to Newton's third law,
|F₁₂| = |F₂₁|
so, F₁₂ = F₂₁ = 0.2N { attractive nature.}
Charge on large sphere is denoted by q₂
force is denoted by F
Given, q₁ = 0.4μC = 0.4 × 10⁻⁶ C
q₂ = 0 8μC = 0.8 × 10⁻⁶ C
F = 0.2 N
∴ use Coulombs law,
F = Kq₁q₂/r²
0.2 = 9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶/r²
r² = (9 × 0.32) × 10⁻³/0.2
r² = 14.4 × 10⁻³
r² = 144 × 10⁻⁴
Taking square root both sides,
r = 12 × 10⁻² m = 12cm
(b) according to Newton's third law,
|F₁₂| = |F₂₁|
so, F₁₂ = F₂₁ = 0.2N { attractive nature.}
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Answer:
Distance between two small sphere is 12cm
Force on 2nd sphere due to 1st sphere = -0.2N
(Attractive force will act between them)
For explanation refer the attachment.
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