32. A particle carrying a charge of 1 micro coulomb (μC) is placed at a distance of 50 cm from a fixed charge where it has a potential energy of 10 J. Calculate (i) the electric potential at the position of the particle (ii) the value of the fixed charge.
Answers
Answer:
Gain in kinetic energy = Difference in potential energy
Potential Energy,
Where q₁ and q₂ are the charges
r₁₂ is the distance between the charges
k is Coulomb's constant ≈ 9×10⁹
Given :
q₁ = +5 μC
q₂ = ±15 μC
r₁₂ = 30
To find: K.E. at the instant q₁ moved 15cm from initial position.
Initial potential energy
(a) Since both charges are positive, the charges repel. Hence the charge q₁ moves away by 15 cm.
Hence r₁₂ = 30+15 =45 cm
K.E. = E - E' = 2.25 - 1.5 = 0.75
(b) Since both charges are opposite, the charges attract. Hence the charge q₁ moves closer by 15 cm.
Hence r₁₂ = 30-15 =15 cm
K.E. = E' - E = 4.5 - 2.25 = 2.25
Answer:
Since both charges are positive, the charges repel. Hence the charge q₁ moves away by 15 cm. (b) Since both charges are opposite, the charges attract. Hence the charge q₁ moves closer by 15 cm
Explanation:
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