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QPage 339: Q2: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q2 Solution: A point charge is moving along a straight line with a constant…

⚠️ Common Exam Traps & Conceptual Pitfalls

When the charge approaches the area, $x$ decreases so $dx/dt=-v$. Forgetting the minus sign gives $d(x^{-2})/dt=-2v/x^3$ and a spurious negative displacement current; use $d(x^{-2})/dt=2v/x^3$ to obtain the correct positive magnitude.

A point charge is moving along a straight line with a constant velocity vv. Consider a small area AA perpendicular to the motion of the charge. Calculate the displacement current through the area when its distance from the charge is xx. The value of xx is not large, so that the electric field at any instant is essentially given by Coulomb's law.

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**Answer:** Id=qvA2πϵ0x3I_d = \frac{q v A}{2\pi \epsilon_0 x^3} **Step-by-Step Explanation:** The electric field at distance xx from a point charge qq is E=q4πϵ0x2E = \frac{q}{4\pi \epsilon_0 x^2}. The electric flux through area AA perpendicular to the motion is ΦE=EA=qA4πϵ0x2\Phi_E = EA = \frac{qA}{4\pi \epsilon_0 x^2}. Displacement current: Id=ϵ0dΦEdt=ϵ0ddt(qA4πϵ0x2)=qA4πddt(x−2)I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \frac{d}{dt}\left(\frac{qA}{4\pi \epsilon_0 x^2}\right) = \frac{qA}{4\pi} \frac{d}{dt}(x^{-2}). Since dx/dt=−vdx/dt = -v (charge approaching), ddt(x−2)=−2x−3dxdt=2vx−3\frac{d}{dt}(x^{-2}) = -2x^{-3} \frac{dx}{dt} = 2v x^{-3}. Thus, Id=qA4π⋅2vx−3=qvA2πx3I_d = \frac{qA}{4\pi} \cdot 2v x^{-3} = \frac{q v A}{2\pi x^3} [citation:3][citation:4].

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