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

     

(4)   

            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

     

(3)    

        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)

     

(4)     

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

 



Q 6 :

Three infinitely long wires with linear charge density λ are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?          [2025]

  • xy + yz + zx = constant

     

  • (x + y)(y + z)(z + x) = constant

     

  • (x2+y2)(y2+z2)(z2+x2) = constant

     

  • xyz = constant

     

(3)

V=E·dr=2kλrdr=2kλ ln r+c

Net potential due to all wires

V=2kλ ln x2+y2+2kλ ln y2+z2+2kλ ln z2+x2+c

for V to be constant

(x2+y2)(y2+z2)(z2+x2)=c'

 (x2+y2)(y2+z2)(z2+x2)=c

where c = constant



Q 7 :

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is:          [2025]

  • 14V

     

  • 25V

     

  • 34V

     

  • V

     

(2)

Using V=E(d)

V = E(10)

VAB=E.4=V10×4=2V5



Q 8 :

The electrostatic potential on the surface of uniformly charged spherical shell of radius R = 10 cm is 120 V. The potential at the centre of shell, at a distance r = 5 cm from centre, and at a distance r = 15 cm from the centre of the shell respectively, are:          [2025]

  • 120 V, 120 V, 80 V

     

  • 40 V, 40 V, 80 V

     

  • 0 V, 0 V, 80 V

     

  • 0 V, 120 V, 40 V

     

(1)

Potential inside shell is equal to potential on surface

Vin=Vsurface=kQR=120 V

at r = 15 cm

V=kQr=120×1015=80 V



Q 9 :

Two metal spheres of radius R and 3R have same surface charge density σ. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes σ1 and σ2, respectively. The ratio σ1σ2 is.          [2025]

  • 19

     

  • 9

     

  • 13

     

  • 3

     

(4)

For conducting sphere, V=σrε0

After contact, V1=V2

σ1r1=σ2r2  σ1σ2=r2r1=3



Q 10 :

The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density λ is              [2023]

  • 2λε0

     

  • λ2ε0

     

  • λ4ε0

     

  • λε0

     

(2)

Potential at centre

V=(λ·πR2)4πε0R2 +(λ·πR1)4πε0R2=λ2ε0



Q 11 :

Two isolated metallic solid spheres of radii R and 2R are charged such that both have same charge density σ. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ', the ratio σ'σ is:                 [2023]

  • 94

     

  • 43

     

  • 53

     

  • 56

     

(4)

Q1'4πε0R=Q2'4πε0(2R)

 Q2'=2Q1'

Q1'+Q2'=Q1+Q2

  Q2'2+Q2'=20πR2σ

  32Q2'=20πR2σ

  Q2'4π(2R)2=23 20πR2σ16πR2

  σ'σ=56



Q 12 :

Which of the following correctly represents the variation of electric potential (V) of a charged spherical conductor of radius (R) with radial distance (r) from the centre?            [2023]

  •  

  •  

  •  

  •  

(3)



Q 13 :

Considering a group of positive charges, which of the following statements is correct?                  [2023]

  • Net potential of the system cannot be zero at a point but net electric field can be zero at that point.

     

  • Net potential of the system at a point can be zero but net electric field can't be zero at that point.

     

  • Both the net potential and the net field can be zero at a point.

     

  • Both the net potential and the net electric field cannot be zero at a point.

     

(1)

V=KQiri

Here, Qi and ri are positive

  V>0



Q 14 :

A point charge 2×10-2C is moved from P to S in a uniform electric field of 30 N C-1 directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0,0,0) m respectively, the work done by electric field will be                 [2023]

  • 1200 mJ

     

  • 600 mJ

     

  • − 600 mJ

     

  • − 1200 mJ

     

(3)

WE=qE·S

       =2×10-2[30i^·(-i^)]

        =2×10-2(-30)=-60×10-2

         =-60100=-0.6 J=-600 mJ



Q 15 :

For a uniformly charged thin spherical shell, the electric potential (V) radially away from the centre (O) of the shell can be graphically represented as         [2023]

  •  

  •  

  •  

  •  

(1)

Vinside=kQR

Voutside=kQr



Q 16 :

Electric potential at a point 'P' due to a point charge of 5×10-9 C is 50 V. The distance of 'P' from the point charge is

(Assume, 14πε0=9×10-9 Nm2C-2)     [2023]

  • 3 cm

     

  • 90 cm

     

  • 9 cm

     

  • 0.9 cm

     

(2)

VP=KQr

50=9×109×5×10-9r

r=4550=910=0.9 m=90 cm



Q 17 :

For a charged spherical ball, electrostatic potential inside the ball varies with r as V=2ar2+b. Here, a and b are constants and r is the distance from the center. The volume charge density inside the ball is -λaε. The value of λ is _________.  (ε=permittivity of the medium)           [2023]



(12)

E=-dVdr=-4arρr3ε0 (compare)

Result inside uniformly charged solid sphere.

ρ=-12aε0

λ=12



Q 18 :

Three concentric spherical metallic shells X, Y and Z of radius a, b and c respectively (a<b<c) have surface charge densities σ, -σ and σ, respectively. The shells X and Z are at the same potential. If the radii of X and Y are 2 cm and 3 cm, respectively, the radius of shell Z is ________ cm.       [2023]



(5)

qx=σ4πa2,  qy=-σ4πb2,  qz=σ4πc2

Potential at x=potential at z

Vx=Vz

qx4πε0a+qy4πε0b+qz4πε0c=qx4πε0c+qy4πε0c+qz4πε0c

σ4πa2a-σ4πb2b+σ4πc2c=4πσ(a2-b2+c2)c

c(a-b+c)=a2-b2+c2

c(a-b)=a2-b2

  c=a+b  c=5 cm



Q 19 :

64 identical drops, each charged up to a potential of 10 mV are combined to form a bigger drop. The potential of the bigger drop will be ______ mV.          [2023]



(160)

Let q=charge on each drop

        V=Kqr                                                ...(i)

Now for combination of 64 drops

      64×43πr3=43πR3

      R=4r and Q=64q

Potential of bigger drop =KQR=K(64q)4r=16Kqr

                                               =16×10 mV=160 mV

Correct answer is 160



Q 20 :

 A point charge of 10-8 C is placed at the origin. The work done in moving a point charge 2μC from point A(4,4,2) to B(2,2,1) is ______J. (14πε0=9×109 in SI units)      [2026]

  • 0

     

  • 15×10-6

     

  • 30×10-6

     

  • 45×10-6

     

(3)

Work done by external agent :

Wext=ΔU;

ΔU Change in potential energy in taking the charge from initial to final configuration

   Wext=14πε0q1q2rf-14πε0q1q2ri

Now,  rf=(2-0)2+(2-0)2+(1-0)2=3 m

             ri=(4-0)2+(4-0)2+(2-0)2=6 m

  Wext=(9×109)×(10-8×2×10-6)[13-16]

                   =3×10-5

                    =30×10-6 J



Q 21 :

There are three co-centric conducting spherical shells A, B and C of radii a, b and c respectively (c>b>a) and they are charged with charges q1, q2 and q3 respectively. The potentials of the spheres A, B and C respectively, are:                   [2026]

  • 14πε0(q1+q2+q3a), 14πε0(q1+q2+q3b), 14πε0(q1+q2+q3c)

     

  • 14πε0(q1a+q2b+q3c), 14πε0(q1+q2b+q3c), 14πε0(q1+q2+q3c)

     

  • 14πε0(q1a+q2b+q3c), 14πε0(q1+q2+q3b), 14πε0(q1+q2+q3c)

     

  • 14πε0(q1+q2+q3a), 14πε0(q1+q2b+q3c), 14πε0(q1a+q2b+q3c)

     

(2)

VA=Kq1a+Kq2b+Kq3c=14πε0(q1a+q2b+q3c)

VB=Kq1b+Kq2b+Kq3c=14πε0(q1+q2b+q3c)

VC=Kq1c+Kq2c+Kq3c=14πε0(q1+q2+q3c)



Q 22 :

The electrostatic potential in a charged spherical region of radius r varies as V=ar3+b, where a and b are constants. The total charge in the sphere of unit radius is α×πaε0. The value of α is _________.  

(Permittivity of vacuum is ε0)                                   [2026]

  • − 12

     

  • − 8

     

  • − 9

     

  • − 6

     

(1)

V=ar3+b

E=-dVdr=-3ar2

ϕclosed=qencε0

qenc=ε0·E·A

=ε0(-3a·(1)2)4π(1)2

=-12πaε0

 x=-12



Q 23 :

Electric field in a region is given by E=Axi^+Byj^, where A=10 V/m2 and B=5 V/m2. If the electric potential at a point (10, 20) is 500 V, then the electric potential at the origin is ________ V.                [2026]

  • 1000

     

  • 0

     

  • 500

     

  • 2000

     

(4)

E=10xi^+5yj^

Vat(10,20)=500 V

ΔV=-E·dr

500-V0=-(0,0)(10,20)(10xi^+5yj^)·(dxi^+dyj^)

500-V0=-[5x2+5y22](0,0)(10,20)

V0-500=(500+5×4002)-(0-0)

V0-500=500+1000

V0=2000 V



Q 24 :

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :   [2026]

  • 32/3:1

     

  • 1:31/3

     

  • 1:22/3

     

  • 1:32/3

     

(4)

Using volume conservation

3(43πr3)=(43πR3)

R=31/3r

ViVf=kqrk(3q)R=R3r=31/3r3r=132/3



Q 25 :

Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure.The electric potential (V) and the electric field (E) at the center O of the pentagon due to these five positive charges are:    [2026]

  • V=5q4πε0r and E=5q4πε0r2r^

     

  • V=5q4πε0r and E=0

     

  • V=0 and E=0

     

  • V=5q4πε0r and E=53q8πε0r2r^

     

(2)

Electric potential V=5kqR

As regular polygon E=0