Physics · Electromagnetic Waves · NEET
Yes, always. In a charging capacitor no real charge crosses the gap, so conduction current there is zero. But the changing electric field creates a displacement current I_d in the gap that is exactly equal to the conduction current I_c in the wires. This keeps the current continuous around the whole circuit. So in numericals, whatever current flows in the wire is the same value that flows as displacement current between the plates.
Use one formula: I_d = C (dV/dt). Multiply the capacitance C (in farad) by the rate of change of voltage dV/dt (in volt per second). Example: C = 20 microfarad and dV/dt = 3 V/s gives I_d = 20e-6 x 3 = 60e-6 A = 60 microampere. Keep units in SI (farad and V/s) so the answer comes out in ampere.
Rearrange the same formula: dV/dt = I_d / C. Example: I_d = 2 mA = 2e-3 A and C = 4 microfarad = 4e-6 F, so dV/dt = 2e-3 / 4e-6 = 500 V/s. Always convert mA to A and microfarad to F before dividing, or the power of ten will be wrong.
Q = CV, so dQ/dt = C (dV/dt). This means the rate of change of charge on the plates is another name for the displacement current: I_d = dQ/dt. If a problem gives dQ/dt directly, that number IS the displacement current, no extra step needed.
Yes, when the problem gives dV/dt or asks you to find it. I_d depends on both C and dV/dt. But if the problem simply gives you the conduction current in the wire and asks for the displacement current, C is not needed, the two currents are equal by continuity.
Differentiate: dV/dt = V0 w cos(wt). Then I_d = C (dV/dt) = V0 w C cos(wt). Note it comes out as cosine (not sine) and its peak value is V0 w C. This is the 2021 NEET question.
A parallel plate capacitor of capacitance 20 microfarad is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires and the displacement current through the plates of the capacitor would be, respectively,
To produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 microfarad, the rate of change of applied variable potential difference (dV/dt) must be
A capacitor of capacitance C is connected across an ac source of voltage V given by V = V0 sin(wt). The displacement current between the plates of the capacitor would then be given by
Try the real previous-year questions from this chapter — each with the answer and a full solution.
I_d = C (dV/dt) = dQ/dt. This is the same as the conduction current in the connecting wires.
Between the plates the electric field changes with time. Maxwell showed this changing field is equivalent to a current called displacement current, I_d = epsilon0 (d(phi_E)/dt), which for a capacitor equals C(dV/dt).
Only when dV/dt or dQ/dt is involved. If you are simply told the wire (conduction) current and asked for the gap current, they are equal, so C is not required.
Use SI: C in farad, dV/dt in V/s, so I_d comes out in ampere. Convert microfarad (1e-6) and mA (1e-3) before substituting.
No. It is not moving charge. It is the effect of a time-changing electric field, but it produces a magnetic field just like a real current and keeps the total current continuous.
