Verified Solution • 2025–26 Session moderate

QPage 339: Q1: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Waves Page 339: Q1 Solution: Show that the dimensions of the displacement current are that…

Show that the dimensions of the displacement current ϵ0dΦEdt\epsilon_0 \frac{d\Phi_E}{dt} are that of an electric current.

Step-by-Step Verified Solution

**Answer:** The dimensions of displacement current are the same as those of electric current. **Step-by-Step Explanation:** Displacement current is defined as Id=ϵ0dΦEdtI_d = \epsilon_0 \frac{d\Phi_E}{dt}. The electric flux ΦE=∫E⃗⋅dA⃗\Phi_E = \int \vec{E} \cdot d\vec{A} has dimensions [E][L2]=[MLT−3A−1][L2]=[ML3T−3A−1][E][L^2] = [MLT^{-3}A^{-1}][L^2] = [ML^3T^{-3}A^{-1}]. Taking the time derivative: [ΦE]/[T]=[ML3T−4A−1][\Phi_E]/[T] = [ML^3T^{-4}A^{-1}]. Multiplying by ϵ0\epsilon_0, which has dimensions [M−1L−3T4A2][M^{-1}L^{-3}T^4A^2]: [Id]=[M−1L−3T4A2]×[ML3T−4A−1]=[A][I_d] = [M^{-1}L^{-3}T^4A^2] \times [ML^3T^{-4}A^{-1}] = [A]. Thus, displacement current has the same dimensions as electric current [citation:3][citation:4].

All Questions in this Chapter

QPage 338: Q1 QPage 338: Q2 QPage 338: Q3 QPage 338: Q4 QPage 338: Q5 QPage 338: Q6 QPage 338: Q7 QPage 339: Q1 QPage 339: Q1 QPage 339: Q2 QPage 339: Q2 QPage 339: Q3 QPage 339: Q3 QPage 339: Q4 QPage 339: Q4 QPage 339: Q5 QPage 339: Q5 QPage 339: Q6 QPage 339: Q7 QPage 339: Q8 QPage 339: Q9 QPages 338-339: Q1 QPages 338-339: Q2 QPages 338-339: Q3 QPages 338-339: Q4 QPages 338-339: Q5 QPages 338-339: Q6 QPages 338-339: Q7 QPages 338-339: Q8 QPages 338-339: Q9

Related Learning Resources & Exam Preparation

Chapter Practice Bank
Practice all topic-wise questions from Electromagnetic Waves
Reference Textbook Solutions
Explore complete chapter exercises for HC Verma: Concepts of Physics - Volume 2
Competitive PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring