Electric Potential Energy Potential And Dipoles(11th And 12th > Physics ) Questions and Answers

Question 1.


Three particles, each of charge 10 μC are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is
Given (14πϵ=9×109 N−m2/C2)


  1.     Zero
  2.     Infinite
  3.     27 J
  4.     100 J
Explanation:-
Answer: Option C. -> 27 J
:
C
For pair of charge U=14πϵ0[10×10−6×10×10−610100+10×10−6×10×10−610100+10×10−6×10×10−610100]
=3×9×109×100×10−12×10010=27 J
Question 2.


Four equal charges Q  are placed at the four corners of a square of each side 'd'. Work done in removing a charge – Q from its centre to infinity is


  1.     0
  2.     √2Q24πϵ0a
  3.     √2Q2πϵ0a
  4.     Q22πϵ0a
Explanation:-
Answer: Option C. -> √2Q2πϵ0a
:
C
Potential at centre O of the square
Four Equal Charges Q  are Placed At The Four Corners Of A ...
V0=4(Q4πϵ0(a/√2))
Work done in shifting (– Q) charge from centre to infinity
W=−Q(V∞−V0)=QV0=4√2Q24πϵ0a=√2Q2πϵ0a
 
Question 3.


There is an electric field E along X-direction. If the work done on moving a charge 0.2 C through a distance of 2m along a line making an angle 60∘ with the X-axis is 4.0, what is the value of E


  1.     √3 N/C
  2.     4 N/C
  3.     5 N/C
  4.     None of these
Explanation:-
Answer: Option D. -> None of these
:
D
W=qV=qE.d
There Is An Electric Field E Along X-direction. If The Work ...
⇒4=0.2×E×(2 cos 60∘)
=0.2E×(2×0.5)
∴E=40.2=20 NC−1
Question 4.


Four identical charges +50 μC each are placed, one at each corner of a square of side 2m. How much external energy is required to bring another charge of +50 μC from infinity to the centre of the square (Given 14πϵ0=9×109 Nm2C2)


  1.     64 J
  2.     41 J
  3.     16 J
  4.     10 J
Explanation:-
Answer: Option A. -> 64 J
:
A
Potential at the centre of square
V=4×(9×109×50×10−6√2)=90√2×104 V
Work done in bringing a charge (q = 50 μC) from ∞ to centre (O) of the square is
W=q(V0−V∞)=qV0
⇒W=50×10−6×90√2×104=64 J
 
Question 5.


Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10 C are connected by a copper wire and then they are separated. Then


  1.     Both the spheres will have the same charge of 10 C
  2.     Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere
  3.     Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere
  4.     Surface charge density on the two spheres will be equal
Explanation:-
Answer: Option C. -> Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere
:
C
After redistribution, charges on them will be different, but they will acquire common potential
i.e. kQ1r1=kQ2r2⇒Q1Q2=r1r2
As σ=Q4πr2⇒σ1σ2=Q1Q2×r22r21⇒σ∝1r
i.e. surface charge density on smaller sphere will be more.
Question 6.


Two charges +3.2×10−19 and −3.2×10−19 C placed  2.4˙A apart form an electric dipole. It is placed in a uniform electric field of intensity 4×105 volt/m. The electric dipole moment is


  1.     15.36×10−29 Coulomb×m
  2.     15.36×10−19 Coulomb×m
  3.     7.68×10−29 Coulomb×m
  4.     7.68×10−19 Coulomb×m
Explanation:-
Answer: Option C. -> 7.68×10−29 Coulomb×m
:
C
Dipole moment p = q (d)
=3.2×10−19×(2.4×10−10)=7.68×10−29C−m
Question 7.


An electron and a proton are at a distance of 1˙A. The dipole moment will be (C×m)    
 


  1.     1.6×1019
  2.     1.6×10−29
  3.     3.2×1019
  4.     3.2×1029
Explanation:-
Answer: Option B. -> 1.6×10−29
:
B
p=q×(d)=1.6×10−19×10−10=1.6×10−29 C−m
    
Question 8.


Point charges q, q, -2q are placed at the corners of an equilateral triangle ABC of side l. The magnitude of net electric dipole moment of the system is


  1.     ql
  2.     2ql
  3.     √3ql
  4.     4ql
Explanation:-
Answer: Option C. -> √3ql
:
C
Point Charges Q, Q, -2q Are Placed At The Corners Of An Equi...
Pnet=√p2+p2+2pp cos60∘=√3p=√3ql (∵p=ql)
 
Question 9.


If an electron starts from rest from a point at which potential is 50 volt to another point at which potential is 70 volt, then its kinetic energy in the final state will be


  1.     3.2×10−10J
     
  2.     3.2×10−18J
  3.     1J
  4.     1 dyne
Explanation:-
Answer: Option B. -> 3.2×10−18J
:
B
K.E.=q0(VA−VB)=1.6×10−19(70−50)=3.2×10−18J
Question 10.


Two point charges each of  +1μC are at a distance of 3 m apart. The work done in bringing the two charges closer to a distance 1 m apart is


  1.     0.006 J
  2.     0.008 J
  3.     0.0006 J
  4.     0.06 J
Explanation:-
Answer: Option A. -> 0.006 J
:
A
W=U2−U1=14πϵ0q1q2[1r2−1r1]=9×109×10−12[11−13]=9×23×10−3W=0.006J
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