NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
358757
To establish an instantaneous displacement current of \(2\;A\) in the space between two parallel plate of \(1 \mu F\) capacitor, the potential difference across the capacitor plates will have to be changed at the rate of
358758
In order to establish an instantaneous displacement current of \(1\;mA\) in the space between the plates of \(2\,\mu F\) parallel plate capacitor, the potential difference need to apply is:
358759
A condenser having circular plates having radius \(2\;cm\) and separated by a distance of \(3\;mm\). It is charged with a current of \(0.1\;A\). The rate at which the potential difference between the plates change is
358760
A parallel plate capacitor with plate area \({A}\) and seperation \({d}\) is charged by a constant current \(i\). Consider a plane surface of area \({A / 4}\), parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?
1 \({i}\)
2 \({2 i}\)
3 \({i / 4}\)
4 \({i / 2}\)
Explanation:
Electric field between the plates of the capacitor is given by \({E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{q}{A \varepsilon_{0}}}\) Flux through the area, \({\phi=\dfrac{q}{A \varepsilon_{0}} \times \dfrac{A}{4}=\dfrac{q}{4 \varepsilon_{0}}}\) Displacement current, \({i_{d}=\varepsilon_{0} \dfrac{d \phi_{E}}{d t}}\) \({=\varepsilon_{0} \times \dfrac{d}{d t}\left(\dfrac{q}{4 \varepsilon_{0}}\right)=\dfrac{i}{4}}\)
358757
To establish an instantaneous displacement current of \(2\;A\) in the space between two parallel plate of \(1 \mu F\) capacitor, the potential difference across the capacitor plates will have to be changed at the rate of
358758
In order to establish an instantaneous displacement current of \(1\;mA\) in the space between the plates of \(2\,\mu F\) parallel plate capacitor, the potential difference need to apply is:
358759
A condenser having circular plates having radius \(2\;cm\) and separated by a distance of \(3\;mm\). It is charged with a current of \(0.1\;A\). The rate at which the potential difference between the plates change is
358760
A parallel plate capacitor with plate area \({A}\) and seperation \({d}\) is charged by a constant current \(i\). Consider a plane surface of area \({A / 4}\), parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?
1 \({i}\)
2 \({2 i}\)
3 \({i / 4}\)
4 \({i / 2}\)
Explanation:
Electric field between the plates of the capacitor is given by \({E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{q}{A \varepsilon_{0}}}\) Flux through the area, \({\phi=\dfrac{q}{A \varepsilon_{0}} \times \dfrac{A}{4}=\dfrac{q}{4 \varepsilon_{0}}}\) Displacement current, \({i_{d}=\varepsilon_{0} \dfrac{d \phi_{E}}{d t}}\) \({=\varepsilon_{0} \times \dfrac{d}{d t}\left(\dfrac{q}{4 \varepsilon_{0}}\right)=\dfrac{i}{4}}\)
358757
To establish an instantaneous displacement current of \(2\;A\) in the space between two parallel plate of \(1 \mu F\) capacitor, the potential difference across the capacitor plates will have to be changed at the rate of
358758
In order to establish an instantaneous displacement current of \(1\;mA\) in the space between the plates of \(2\,\mu F\) parallel plate capacitor, the potential difference need to apply is:
358759
A condenser having circular plates having radius \(2\;cm\) and separated by a distance of \(3\;mm\). It is charged with a current of \(0.1\;A\). The rate at which the potential difference between the plates change is
358760
A parallel plate capacitor with plate area \({A}\) and seperation \({d}\) is charged by a constant current \(i\). Consider a plane surface of area \({A / 4}\), parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?
1 \({i}\)
2 \({2 i}\)
3 \({i / 4}\)
4 \({i / 2}\)
Explanation:
Electric field between the plates of the capacitor is given by \({E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{q}{A \varepsilon_{0}}}\) Flux through the area, \({\phi=\dfrac{q}{A \varepsilon_{0}} \times \dfrac{A}{4}=\dfrac{q}{4 \varepsilon_{0}}}\) Displacement current, \({i_{d}=\varepsilon_{0} \dfrac{d \phi_{E}}{d t}}\) \({=\varepsilon_{0} \times \dfrac{d}{d t}\left(\dfrac{q}{4 \varepsilon_{0}}\right)=\dfrac{i}{4}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI15:WAVES
358757
To establish an instantaneous displacement current of \(2\;A\) in the space between two parallel plate of \(1 \mu F\) capacitor, the potential difference across the capacitor plates will have to be changed at the rate of
358758
In order to establish an instantaneous displacement current of \(1\;mA\) in the space between the plates of \(2\,\mu F\) parallel plate capacitor, the potential difference need to apply is:
358759
A condenser having circular plates having radius \(2\;cm\) and separated by a distance of \(3\;mm\). It is charged with a current of \(0.1\;A\). The rate at which the potential difference between the plates change is
358760
A parallel plate capacitor with plate area \({A}\) and seperation \({d}\) is charged by a constant current \(i\). Consider a plane surface of area \({A / 4}\), parallel to the plates, and drawn symetrically between the plates. What is the displacement current through this area?
1 \({i}\)
2 \({2 i}\)
3 \({i / 4}\)
4 \({i / 2}\)
Explanation:
Electric field between the plates of the capacitor is given by \({E=\dfrac{\sigma}{\varepsilon_{0}}=\dfrac{q}{A \varepsilon_{0}}}\) Flux through the area, \({\phi=\dfrac{q}{A \varepsilon_{0}} \times \dfrac{A}{4}=\dfrac{q}{4 \varepsilon_{0}}}\) Displacement current, \({i_{d}=\varepsilon_{0} \dfrac{d \phi_{E}}{d t}}\) \({=\varepsilon_{0} \times \dfrac{d}{d t}\left(\dfrac{q}{4 \varepsilon_{0}}\right)=\dfrac{i}{4}}\)