A point charge 20 µC is shifted from infinity to a point P in an electric field with zero acceleration. If the potential of that point is 1000 volt, then
(i) Find out work done by external agent against electric field?
(ii) What is the work done by electric field?
(iii) If the kinetic energy of charge particle is found to increase by 10 mJ when it is brought from infinity to point P, then what is the total work done by external agent?
(iv) What is the work done by electric field in the part (iii)
(v) If a point charge 30 C is released at rest at point P, then find out its kinetic energy at a large distance?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
[(i) q( Δ V) = 20 mJ.
(ii)– 20 mJ
(iii) q( Δ V) + Δ K.E. = 30 mJ
(iv) –20mJ
(v) 30 mJ]
Sol. (speed) v = const
∞ (q)-------------------------(q) P
V ∞ = 0 V P = 1000 V
(i) W ext = q (V P – V ∞ ) = 20 × 10 –6 × 1000 = 20 mJ
(ii) W elec = q (V ∞ – V P ) = – 20 mJ = – Δ U
(iii) W ext = Δ U + Δ k = q Δ V + Δ K = 20 + 10 = 30 mJ
(iv) W elec = q (V ∞ – V P ) = – 20 mJ = – Δ U
(v) P (q) V = 0 --------------------(q) V = (k.E = ?)
V P = 1000 V ∞
w elec = Δ K since no external force.
∴ W elec . = – Δ U = – q Δ V = – q ( V ∞ – V p ) = – 30 × 10 –6 (0 – 1000) = 30 mJ = Δ K = K ∞ – 0
∴ K ∞ = 30 mJ
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