A parallel plate capacitor consists of two circular plates each of radius $2 \,cm$, separated by a distance of $0.1 \,mm$. If the potential difference across the plates is varying at the rate of $5 \times 10^{6} Vs ^{-1}$, then the value of displacement current is
According to the question, a parallel plate capacitor is shown in the figure below,
Given, radius $=2\, cm$, distance between plates,
$d=0.1\, mm$ and $\frac{d V}{d t}=5 \times 10^{6}\, Vs ^{-1}$
Displacement current in the capacitor,
$I_{d}=\varepsilon_{0} \frac{d \phi}{d t}=\varepsilon_{0} \frac{d E}{d t} A $
$I_{d}=\varepsilon_{0} A \frac{d E}{d t}$
$\Rightarrow\, I_{d}=\varepsilon_{0} \frac{A}{d} \frac{d V}{d t} $
$\left(\because \,E=\frac{V}{d}\right)$
Putting the given values, we get
$\Rightarrow\, I_{d}=\frac{8.85 \times 10^{12} \times \pi \times\left(4 \times 10^{4}\right) \times 5 \times 10^{6}}{1 \times 10^{-4}} A$
$\Rightarrow\, I_{d}=556 \cdot 28 \times 10^{-6} \,A$
$ =0.556 \,mA$
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Top Questions on electrostatic potential and capacitance
The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.
Some major things that we should know about electric potential:
They are denoted by V and are a scalar quantity.
It is measured in volts.
Capacitance
The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.
The capacitor is in Series and in Parallel as defined below;
In Series
Both the Capacitors C1 and C2 can easily get connected in series. When the capacitors are connected in series then the total capacitance that is Ctotal is less than any one of the capacitor’s capacitance.
In Parallel
Both Capacitor C1 and C2 are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is Ctotal is any one of the capacitor’s capacitance.
