NEET Physics Electrostatic Potential and Capacitance Questions Solved


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(2)

VP=V1+V2+V3+V4     = Kq5a2+Kqa2+Kqa2+Kq5a2     = 4Kq5a+4Kqa

V=2Kqa2+25

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(2)

                     

Electric filed due to +Q=σ20                                          =Q20AElectric field due to -Q=Q20ANet E.F. =Q0A=E Electric field due to +Q on -Q= Q20A=E2 Force on -Q due to +Q=-QE2

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(2)

Initial charge on the spheres:

Q1=C1V1             &            Q2=C2V2Initial potential energy of the spheres-U1=12C1V12                          U2=12C2V22Charges will flow between the spheres until they have same electric potential.Let q charge flows from sphere 1 to 2. Then,                          V1'=V2'                              Q1-qC1=Q2+qC2                      q=Q1C2-Q2C1C1+C2 = C1V1C2-C2V2C1C1+C2                      q=C1C2C1+C2V1-V2                      V1'=V2'=1C1Q1-C1C2C1+C2V1-V2                                       =1C1C1V1C1+C2-C1C2V1-V2C1+C2                                       =1C1C12V1-C1C2V2C1+C2                                       =C1V1-C2V2C1+C2Final Potential enrergy of the spheres-U1'=12C1V1'2=12C1C1V1-C2V2C1+C22U2'=12C2V2'2=12C2C1V1-C2V2C1+C22

Heat produced = Loss in potential energy                           =Final potential enrgy - initial potential enrgy = Uf-Ui                           =U1'+U2'-U1+U2                           =12C1+C2C1V1-C2V2C1+C22-12C1V12+C2V22                           =-12C1V1-C2V22-C1+C2C1V12+C2V22C1+C2                           =-12C12V12+C22V22-2C1V1C2V2-C12V12-C1C2V22-C1C2V12-C22V22C1+C2                           =12C1+C2C1C2V12+C1C2V22-2C1C2V1V2                           =C1C22C1+C2V1-V22

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The electric potential at the surface of a charged solid sphere of insulator is 20V. The value of electric potential at its centre will be

1.  30V

2.  20V

3.  40V

4.  Zero

(1)

For a solid charged non-conducting sphere, for an inside point at distance r from center

V=kQ2R3(3R2-r2)At center, r=0Vcenter=3kQ2R

At surface r=R, Vs=kQR Vcenter=3Vs2                   =32X20 = 30V

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