Id=ϵ0dtdϕe=ϵ0dtd(EAcos0∘)Id=ϵ0AdtdE∵E=ϵ0σ;Id=ϵ0Adtd(ϵ0σ)⇒Id=A(dtdσ)= constant Also B due to Id is max m at surface of imaginary cylinder 
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is :
Held on 30 Apr 2025 · Verified 9 Jul 2026.
zero at all places
constant between the plates and zero outside the plates
non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
zero between the plates and non-zero outside
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :- 
The electric potential at distance r from a point charge q is V = q/(4πε₀r). The electric field E is:
To an ac power supply of 220 V at 50 Hz , a resistor of $20 \Omega$, a capacitor of reactance $25 \Omega$ and an inductor of reactance $45 \Omega$ are connected is series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively-
A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is :
Work through every NEET UG Electromagnetism PYQ, year by year.