CBSE Class 12 Physics Electrostatic Potential and Capacitance Worksheet Set A

CBSE Class 12 Physics Electrostatic Potential and Capacitance Worksheet Set A

TWO MARKS QUESTIONS

Question. Four point charges 1μC, 1μC, -1μC and -1 μC are placed at corners of a square of each side 0.1 m (i) calculate electric potential at centre O of square (ii) if E is midpoint of BC, what is work done in carrying an electron from O to E?
Answer : (0,0)

Question.Write an expression for potential energy of two charges q1 and q2 at r1 and r2 in a uniform electric field E.
Answer : [q1 V(r1) + q2 V(r2) + q1q2/4πε0 (r1 – r2)] 

Question. Net capacitance of three identical capacitors in series is 3 μF. What is their net capacitance if connected in parallel?
Find the ratio of the energy stored in the two configurations if they are both connected to the same source.
Answer : . Cs=9μF as they are connected in series.
When connected in parallel Cp= 27 μF
Since voltage is same, so Energy is directly proportional to capacitance
Es/Cs= Ep/Cp
So, Es/Ep = Cs/Cp
Es:Ep= 1:3

Question. Plot a graph showing the variation of Coulomb force(F) versus (1/r2) where r is the distance between the two charges of each pair of charges (1 μC, 2μC) and (2 μC, -3μC). interpret the graphs obtained.
Answer : (Image)

Question. A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field as shown in figure. (i) calculate the potential difference between A and C. (ii) at which point ( of the two) is the electric potential more and why?
Answer : (i) since electric field is conservative in nature, the amount of work done depends on initial and final position.
So work done W=F.d = q E.d = -qE4cos 1800= +4qE
Hence VC-VA = W/q
Hence VC-VA = W/q = 4E

Question. Two charged spherical conductors of radii R1 and R2 when connected by a connecting wire acquire charges q1 and q2 respectively. Find the ratio of their charge densities in the terms of their radii?
Answer :  (on connecting potential will be same. So q1/ 4πε0R1 = q2/4πε0R2 so q1/q2 = R1/R2. And the charge density ratio= σ1:σ2= q1/4πR12: q2/4πR22 so ratio of their charge densities R2/R1)

Question. Two point charges 4Q and Q are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero?
Also calculate the electrostatic potential energy the system of charges, taking the value of Q= -2x10-7 C.
Answer : Let a point inbetween at the distance x from 4 Q then find electric field intensity due to both of the charges. On solving X=66.7 m or x=2 m where x is the distance of point p from charge 4Q where electric field intensity is zero.
Electrostatic potential energy U=1/4π ε0 . q1q2/r= 41.44x10-3 J    

Question. An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 1800.
Answer : (i) The dipole moment of dipole is P= qx (2a)
Force on –q at A = -qE
force on +q at B = +qE
net force on the dipole = qE –qE = 0
(ii) Work done on dipole
W = ∆U = PE( cosɵ1 - cosɵ2)
= PE( cos0o – cos180o)
W = 2PE

THREE MARKS QUESTIONS

Question. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (i) a charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (ii) is the electric field intensity inside a cavity with no charge is zero, even if the shell is not spherical? Explain.
Answer : ( -q/4πr12, Q+q/4πr22,0 ) (when charge q is placed at the centre of the shell, a charge –q is induced at the inner surface and +q at the outer surface of the shell)

Question. Define electric flux. Is it a scalar or a vector quantity? A point charges q is at a distance of d/2 directly above the centre of a square of side d. Use Gauss' law to obtain the expression for the electric flux through the square. (b) If the point charge is now moved to a distance 'd' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.
Answer : (a) Electric flux is defined as the number of electric filed lines crossing the per unit area. It is scalar quantity. When cube is of side d and point charge q is at the center of the cube then the total electric flux due to this charge will pass evenly through the six faces of the cube. So, the electric flux through one face will be equal to 1/6 of the total electric flux due to this charge. i.e., Flux through 6 faces = q/ 0 Flux through 1 face = 1/6 x q/ 0 (b) If we moved point charge d from centre and square side changes to 2d, still the point charge can be imagined at the center of a cube of side 2d. Again the flux through one face of the cube will be 1/6 of the total electric flux due to the charge q. Hence, the electric flux through the square will not change and it will remain the same i.e., q/6∈o.

Question. A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge Q/2 is placed at the centre C and another charge +2Q is placed outside the shell at A at a distance x from the centre as shown in the figure.
(i) Find the electric flux through the shell
(ii) State the law used
(iii) Find the force on the charges at the centre C of the shell and at the point A.
Answer : (i) total charge enclosed = Q+Q/2 = 3Q/2
So, Ф = 3Q/2ε0
ii) Statement of Gauss’s law along with mathematical expression
(iii) Force on charge at A, F = 3Q2/4πε0x2

Question. Two identical parallel plate capacitors A and B are connected to a battery of V volts with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
Answer : Two capacitors are connected in parallel. Hence, the potential on each of them remains the same. So the charge on each capacitor is
QA =QB = CV
Net capacitance with switch S closed = C+C =2C so, energy stored = 1/2X2CXV2 =CV2
After the switch S is opened, capacitance of each capacitor= KC
In this case, voltage only across A remains the same.
The voltage across B changes to V’=Q/C=Q/KC
Energy stored in capacitor A= 1/2KCV2
Energy stored in capacitor B= Q2/2KC
Total energy stored=1/2KCV2+ Q2/2KC= CV2[K2+1]/2K
Required ratio= 2K/(K2+1)

Question. A positive point charge (+q) is kept in the vicinity of an uncharged conducting plate. Sketch the electric field lines originating from the point on to surface of the plate.
Derive expression for the electric field at the surface of a charged conductor.
Answer : (Image)

Question. Use Gausses’ law to derive the expression for the electric field between two uniformly charged parallel sheets with surface charge densities σ and –σ respectively.
Answer : Let us consider two uniformly charged parallel sheets carrying charge densities +σ and –σ respectively, separated by a distance r
By Gauss’ s law it is known that electric field due to a uniformly charged infinite plane sheet at any nearby point is E=σ/2ε0.
The electric field is directed normally outward from the plane sheet for positive charge and normally inward for negative charge.
Let r represents unit vector directed from positive plate to negative plete.
Now electric field intensity at any point P between the plates is given by
(i) E1=+σ/ 2ε0r (due to positive charge)
(ii) E2=- σ/ 2ε0r (due to negative charge)
So , electric field intensity at point P is given by
E= E1+ E2 = σ/ 2ε0r + σ/ 2ε0r = σ/ ε0r

Question. The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between plates. What will be the effect on its capacitance, charge, potential difference, electric field, energy stored? Justify your answer.
Answer : As the battery is connected so p.d. will be same across the plates of the capacitor . V= same, C0= ε0A/d but on inserting the slab C= K ε0A/d so C= KC0, Q= CV= KQ0, E will be same. U= ½ CV2= KU0

Question. The battery connected to a parallel plate capacitor is removed and a dielectric slab is inserted between plates. What will be the effect on its capacitance, charge, potential difference, electric field, energy stored? Justify your answer. Where does the loss of energy stored go?
Answer : As battery is removed so charge will remain same as the source of charge has been removed. Explain further.
- (q same, kC0, V0/k, E0/k, U0/k). in this process dielectric slab is polarized. The energy lost is spent in polarizing the slab)

FIVE MARKS QUESTIONS

Question. a. Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
b. Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.
Answer : . It is defined as the product of either charge and the length of the dipole.
Mathematically, P= q (2a).
It is a vector quantity.

Question. Use Gauss’ Law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.
a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
b) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2>r1)
Answer :  

Question. a. Deduce the expression for the potential energy of an electric dipole of dipole moment P placed in a uniform electric field E.
Find out orientation of the dipole when it is in (i) stable equilibrium (ii) unstable equilibrium.
b. Figure shows a configuration of the charge array of two dipoles

Obtain the expression for the dependence of potential r for r˃>a for a point p on the axis of this array of charges.
Answer : a. Let us suppose that the electric dipole is brought from infinity in the region of a uniform electric field such that the electric dipole moment P always remains perpendicular to electric field. The electric forces on charges +q and -q are +qE and –qE, along the field direction and opposite to the field direction respectively.
As charges +q and –q traverse equal distance under equal and opposite forces; therefore net work done (W1) in bringing the dipole in the region of electric field perpendicular to the field direction will be zero.
Now, the dipole is rotated and brought to orientation making an angle ɵ with the field direction
(ɵ₀ = 90₀ and ɵ1 =ɵo), therefore work done
W2 = PE( cos ɵ₀- cos ɵ1) = -PEcosɵ
So total work done in bringing the dipole from infinity . i.e. electric potential energy of the dipole
U = W1+W2 = 0- PEcosɵ = -PEcosɵ
In vector form U = - P.E
When ɵ=0o, i.e. the dipole is along the direction of the electric field corresponds to stable equilibrium.
When ɵ =1800 i.e. the dipole is anti parallel to the direction of the electric field corresponds to unstable equilibrium.
b. potential at the point P is given by
V = 1/4πɛ₀[q/(r+a) +(-2q)/r +q/r-a]
= q/4πɛ₀[1/(r+a) -2/r+ 1/(r-a)]
= q/4πɛ₀[r2+a2/r(r2-a2)]
As r>> a , V = q/4πɛ₀r

Question. a. Define electric flux. Write its S.I unit.
b.Using Gauss’ Law, obtain the electric flux due to a point charge ‘q’ enclosed in a cube of side ‘a’.
c. Show that the electric field due to a uniformly charged plane sheet at any point distant x from it, is independent of x.
Answer : (a) The number of electric field lines passing through an area normally is called electric flux.
Its SI unit is Nm2C-1 or JmC-1 or Vm.
b. According to Gauss’ slaw states that the total electric flux through a closed surface is equal to 1/ε0 times the magnitude of the charge.
i.e. Ф =q/ε0
Here the charge enclosed by the cube is q , so electric flux through the cube is given by
Ф =q/ε0

              Very Short Answer

Q1)       Who developed first voltaic pile or battery?

Q2)       Define electrostatic potential?

Q3)       Give some example of conservative forces?

Q4)       What kind of charges consists of electric dipole?

Q5)       Define potential energy of charge?

              Short Answer

Q6)       What are the contrasting features of the electric potential dipole?

Q7)       What do you mean by the potential due to the system of charges?

Q8)       Define electrostatic force?

Q9)       What is potential due to an eclectic dipole?

Q10)     What are the two important points state the relation between electric field and potential?

              Long Answer

Q11)     Explain electrostatics of conductors?

Q12)     State the difference between electric field and gravitational field?

Q13)     Write short note on equipotential surfaces?

Q14)     State the difference between polar dielectrics and non-polar dielectrics?

Q15)     Write short on electrostatic shielding?