11) Two point charges QA = 10 micro c and Q8 = - 4 micro c separated by a distance of
50cm.Calculate the force between them. Suppose the spheres A and B have identical
sizes.
A third sphere of the same size but uncharged is brought in contact with the first, then
brought in contact with the second, and finally removed from both. What is the new
force of repulsion between A and B? [3]
Answers
Answer:
36/25
Explanation:
in first contact both body get half charge and after second contact charge equally distributed so final charge on second is 3q/4so final force is f/8
attractive force of 1.44 N Earlier and later on 0.09 N repulsive Force
Explanation:
F = k QA * QB / r²
QA = 10 μ C = 10 * 10⁻⁶ C
QB = - 4 μ C = -4 * 10⁻⁶ C
k = 9×10⁹ N⋅m²⋅C⁻²
r = 50 cm = 0.5 m
F = 9×10⁹ * (10 * 10⁻⁶)*( -4 * 10⁻⁶ ) / (0.5)²
= - 36/25
= -1.44 N
attractive force of 1.44 N
A third sphere of the same size but uncharged is brought in contact with the QA = 10 μ C
=> Charge on both = 5 μ C
Now 5 μ C in Contact with - 4 μ C
Hence charge on both = 0.5 μ C
QA = 5 μ C
QB = 0.5 μ C
F = 9×10⁹ * (5 * 10⁻⁶)*( 0.5 * 10⁻⁶ ) / (0.5)²
F = 9/100
F = 0.09 N repulsive Force
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