Electrostatics MCQs for NEET — Physics Questions with Answers

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64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be

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Explanation

$ V = { 64q \over 4 \pi \varepsilon R } = { 64q \over 4 \pi \varepsilon_0 (4r) } $ $ C = A \pi \varepsilon_0 / d $

Two metal plate form a parallel plate capacitor. The distance between the plates is d. A metal sheet of thickness d/2 and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases ?

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Explanation

$ C^1 = { A \varepsilon_0 \over d-t(1 - {1 \over k })}$ use this equation

A parallel plate capacitor has plate of area A and separation d. It is charged to a potential difference $V_o$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

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Explanation

Work done = Final energy - Initial energy = $ { Q^2 \over 2C^1} - { Q^2 \over 2C } $

Two identical capacitors have the same capacitance C. one of them is charged to a potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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Explanation

$U_i = { 1 \over 2 } C ( V_1^2 + V_2^2 ) $ and $ V = { q_1 + q_2 \over c_1 + c_2 } ={V_1 + V_2 \over 2 } $ $ U_f = { 1 \over 2 } (2C) V^2 Now find U_i U_f $

A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

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Explanation

$C_1 = { A \varepsilon_0 \over d/10 } $ $C_2 = { A \varepsilon_0 \over d/2 } \Rightarrow C_5 = { 5C \over 3 } $ $ \therefore Percentage increase in capacitance = { C_5 - C \over C } \times 100 \% $

The capacities of three capacitors are in the ratio 1 : 2 : 3. Their equivalent capacity when connected in parallel is 60/11 F .more then that when they are connected in series. The individual capacitors are of capacities in $ \mu F$

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Explanation

Theory base question

Electric field is given by E =100x2 . Find the potential difference between x= 10 and x= 20 m. [This question is only for Dropper and XII batch]

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Explanation

E= -dVdx      dV= -Edx      ABdV         =-ABE.dx      VB-VA  =-1020100x2 =-5 voltsPotential difference= 5 volt.

The electric field vector in a region given by E=(3i^+4yj^)Vm-1. Calculate the potential at (1m, 1m) taking potential at origin to be zero. [This question is only for Dropper and XII batch]

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Explanation

Here, E=i^+4yj^, r= 0i^+0j^, r2= i^+j^, dr= dxi^+dyj^, V1=0, V2=V

 V-0=-T1T2(3i^+4yj^).(dxi^+dyj^)             V-0=-013dx -  014ydy          V=-3-4y2201  =-3-42  =-5V

Each of a parallel-plate air capacitor has an area S. What amount of work has to be performed to slowly increase the distance between the plates from x1 to x2. If the voltage across the capacitor, which is equal to V, is kept constant in the process. [This question is only for Dropper and XII batch]

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Explanation

When voltage is kept constant, the force acting on each plate of capacitor will depend on the distance between the plates.

So, elementary work done by agent, in its displacement over a distance dx, relative to the other,
dW=-Fxdx

But, Fx=-σ(x)2ε0Sσ(x) and σ(x)=ε0Vx

Hence, W= dW= x1x212ε0SV2x2dx=ε0SV221x1-1x2

Using the concept of energy density, find the total energy stored by a shell of radius R and charge Q. [This question is only for Dropper and XII batch]

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