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Two hollow concentric spheres A and B are enclosing electric charges 2 C and 8 C respectively. If the radius of A is less than that of B, the ratio of electric flux through A to the electric flux through B is

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Important Questions on Electrostatics

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Consider a region in free space bounded by the surfaces of an imaginary cube having sides of length a as shown in the figure. A charge +Q is placed at the centre O of the cube. P is such a point outside the cube that the line OPperpendicularly intersects the surface ABCD at R and also OR=RP=a/2. A charge +Q is placed at point P also. What is the total electric flux through the five faces of the cube other than ABCD?

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During charging a capacitor, variations of potential V of the capacitor with time t is shown as,
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If the plates of a parallel plate capacitor are not equal in area, then quantity of charge
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Two isolated charged metallic spheres of radii R1 and R2 having charges Q1 and Q2, respectively, are connected to each other, then there is:

 

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A parallel plate capacitor of capacitance C is as shown. A thin metal plate A is placed between the plates of the given capacitor in such a way that its edges touch the two plates as shown. The capacity a cross P and Q now becomes

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A parallel plate capacitor with plate area 6 cm2 and a plate separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant K1=10, K2=12 and K3=14. The dielectric constant of a material which when fully inserted in the above capacitor, gives same capacitance would be:

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The plates of a parallel plate capacitor are charged upto 200 V. A dielectric slab of thickness 4 mm is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of the dielectric slab is
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Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then: