A parallel plate capacitor of capacitance 20 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The charge $Q$ on a capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the potential difference.The conduction current $i_c$

Vedclass · Accepted Answer

$60 \; \mu A, 60 \; \mu A$

Vedclass · Answer

(B) The charge $Q$ on a capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the potential difference.The conduction current $i_c$ in the connecting wires is the rate of change of charge: $i_c = \frac{dQ}{dt} = C \frac{dV}{dt}$.Given $C = 20 \; \mu F$ and $\frac{dV}{dt} = 3 \; V/s$, we have $i_c = 20 \; \mu F 	imes 3 \; V/s = 60 \; \mu A$.According to Maxwell's modification of Ampere's law, the displacement current $i_d$ between the plates of a capacitor is equal to the conduction current in the connecting wires.Therefore, $i_d = i_c = 60 \; \mu A$.Thus, the conduction current is $60 \; \mu A$ and the displacement current is $60 \; \mu A$.

$A$ parallel plate capacitor of capacitance $20 \; \mu F$ 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:

Similar Questions

The capacitance of a capacitor is $2 \ pF$. If the electric field inside the capacitor changes at a rate of $10^{12} \ V \ s^{-1}$,the displacement current is ..... $A$.

$A$ magnetic field can be produced by

The Maxwell-modified form of Ampere's circuital law is .......

According to Maxwell,a changing electric field produces .......

$A$ parallel plate capacitor is charged to $60\,\mu C$. Due to a radioactive source,the plate loses charge at the rate of $1.8 \times 10^{-8}\, C/s$. The magnitude of the displacement current is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module