Question 1:
easyTo produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance 4 \(\mu\)F, the rate of change of applied variable potential difference \(\left(\frac{dV}{dt}\right)\) must be
Displacement current is given by \(I_d = C \frac{dV}{dt}\). Substituting the values: \(2 \times 10^{-3} = 4 \times 10^{-6} \frac{dV}{dt} ⇒ \frac{dV}{dt} = 500\text{ V/s}\).
Question 2:
easyDisplacement current is same as:
Displacement current represents the rate of change of electric displacement field and is not caused by real movement of charges like conduction current.
Question 3:
easyThe potential difference between the plates of a parallel plate capacitor is changing at the rate of \(10^6\text{ V/s}\). If the capacitance is \(2\ \mu\text{F}\), the displacement current in the dielectric of the capacitor will be:
Displacement current is given by \(I_d = C \frac{dV}{dt}\). Substituting \(C = 2 \times 10^{-6}\text{ F}\) and \(\frac{dV}{dt} = 10^6\text{ V/s}\), we find \(I_d = 2\text{ A}\).
Question 4:
easyModified ampere circuital law is given by (symbols have their usual meaning)
The generalized Ampere's circuital law (or Ampere-Maxwell law) includes both conduction current \(I_C\) and displacement current \(I_D\) as sources of magnetic fields, expressed as \(\oint \vec{B} \cdot d\vec{l} = \mu_0(I_C + I_D)\).
Question 5:
easyA capacitor of capacitance \(C\), is connected across an ac source of voltage \(V\), given by \(V = V_0\sin\omega t\). The displacement current between the plates of the capacitor, would then be given by
The displacement current is equal to the conduction current, which is \(I_d = \frac{dq}{dt}\). Since \(q = CV = C V_0 \sin\omega t\), differentiating with respect to time gives \(I_d = V_0 \omega C \cos\omega t\).
Question 6:
easyDisplacement current exists within the gap between the plates of capacitor when the electric field between its plates
Displacement current is given by \(I_d = \varepsilon_0 \frac{d\Phi_E}{dt}\). It exists whenever there is a changing electric field, whether it increases or decreases with time.
Question 7:
easyDisplacement current exists within the gap between the plates of capacitor when the electric field between its plates
Displacement current is given by \(I_d = \varepsilon_0 \frac{d\Phi_E}{dt}\). Hence, it exists whenever the electric field changes (increases or decreases) with time.
Question 8:
easyAssertion (A): Conduction and displacement current may be present in the same region of space.
Reason (R): There is no perfectly conducting or perfectly insulating medium.
In a real dielectric medium with some conductivity, both conduction current (due to charge movement) and displacement current (due to changing electric fields) can exist simultaneously. As no material is perfectly conducting or insulating, this coexistence is possible.
Question 9:
easyAssertion (A): A magnetic needle when placed in between the plates of a parallel plate capacitor under charging, the needle shows deflection.
Reason (R):As the charge on the capacitor plates increases, the electric field and the electric flux between the plates changes which generates a magnetic field.
During capacitor charging, the changing electric field produces a displacement current \(I_d = epsilon_0 frac{dPhi_E}{dt}\). This displacement current generates a magnetic field, causing the needle to deflect. Thus, A and R are true, and R explains A.
Question 10:
moderateTo produce an instantaneous displacement current of 2 mA in the space between the parallel plates of a capacitor of capacitance \(4 \mu\text{F}\), the rate of change of applied variable potential difference \(\left(\frac{dV}{dt}\right)\) must be
Since displacement current is given by \(I_d = C \frac{dV}{dt}\), we find \(\frac{dV}{dt} = \frac{I_d}{C} = \frac{2 \times 10^{-3} \text{A}}{4 \times 10^{-6} \text{F}} = 500 \text{V/s}\).