Class 12 Physics - Chapter Electrostatic Potential and Capacitance NCERT Solutions | Show that the force on each plate of a p
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx.
Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density
A = Area of each plate
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
Electric intensity is given by,
However, capacitance,
Charge on the capacitor is given by,
Q = CV
The physical origin of the factor,
, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value,
, of the field that contributes to the force.
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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10-12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
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Answer the following questions regarding earth's magnetism:
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Would you expect a greater or smaller dip angle in Britain?
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(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
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(f ) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?
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Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.
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