Physics-
General
Easy
Question
Two spherical conductors each of capacity are charged to potentials and . These are then connected by means of a fine wire. The loss of energy will be
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The correct answer is:
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The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate )
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The diameter of each plate of an air capacitor is . To make the capacity of this plate capacitor equal to that of diameter sphere, the distance between the plates will be
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y= , then at x = e is
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lf ST and SN are the Iengths of the subtangent and the subnormal at the point on the curve , where ], then
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The normal to the curves and at angle always passes through the fixed point
The normal to the curves and at angle always passes through the fixed point
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If =and = then
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The distance between the circular plates of a parallel plate condenser in diameter, in order to have same capacity as a sphere of radius is
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A frictionless dielectric plate is kept on a frictionless table . A charged parallel plate capacitance (of which the plates are frictionless) is kept near it. The plate is between the plates. When the plate is left between the plates
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As shown in the figure, a very thin sheet of aluminium is placed in between the plates of the condenser. Then the capacity
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Force of attraction between the plates of a parallel plate capacitor is
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A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is. If a glass plate (dielectric constant) of is introduced between them, then the capacity will become
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drops each having the capacity and potential are combined to form a big drop. If the charge on the small drop is, then the charge on the big drop will be
drops each having the capacity and potential are combined to form a big drop. If the charge on the small drop is, then the charge on the big drop will be
Physics-General
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Separation between the plates of a parallel plate capacitor is and the area of each plate is . When a slab of material of dielectric constantand thickness is introduced between the plates, its capacitance becomes
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The energy of a charged capacitor is given by the expression ( = charge on the conductor and C = its capacity)
The energy of a charged capacitor is given by the expression ( = charge on the conductor and C = its capacity)
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