Q 1 :    

Two charged conducting spheres of radii a and b are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is              [2024]

  • ba

     

  • ab

     

  • ab

     

  • ab

     

(D)      Va=Vb

             KQaa=KQbbQaQb=ab

 



Q 2 :    

An electric charge 10-6μC is placed at origin (0, 0) m of X-Y co-ordinate system. Two points P and Q are situated at (3,3)m and (6,0)m respectively. The potential difference between the points P and Q will be                                    [2024]

  • 3V

     

  • 6V

     

  • 0V

     

  • 3V

     

(C)    Potential difference =KQr1-KQr2

          r1=(3)2+(3)2

          r2=(6)2+0

          As r1=r2=6m

          So potential difference = 0.

 



Q 3 :    

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

 

Assertion (A): Work done by electric field on moving a positive charge on an equipotential surface is always zero.

 

Reason (R): Electric lines of forces are always perpendicular to equipotential surfaces.

 

In the light of the above statements, choose the most appropriate answer from the given below:                                [2024]

  • Both (A) and (R) are correct but (R) is not the correct explanation of (A)

     

  • (A) is correct but (R) is not correct

     

  • (A) is not correct but (R) is correct

     

  • Both (A) and (R) are correct and (R) is the correct explanation of (A)

     

(D)      Electric line of force are always perpendicular to equipotential surface so angle between force and displacement will always be 90o. So work done is equal to 0.

 



Q 4 :    

At the centre of a half ring of radius R = 10 cm and linear charge density 4 n Cm-1, the potential is xπ V. The value of x is _______ .             [2024]



(36)         Potential at centre of half ring

                 V=KQR=KλπRR

                  V=KλπV=9×109×4×10-9π

                      V=36πV

 



Q 5 :    

The electric potential at the surface of an atomic nucleus (z=50) of radius 9×10-13 cm is ______ ×106V              [2024]



(8)     Potential = kQR=k.ZeR

          =9×109×50×1.6×10-199×10-13×10-2=8×106V