(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.


i) Electric field due to a uniformly charged infinite plane sheet:


Consider an infinite thin plane sheet of positive charge with a uniform charge density straight sigma on both sides of the sheet. Let a point be at a distance a from the sheet at which the electric field is required.

The gaussian cylinder is of area of cross section A.

Electric flux crossing the gaussian surface,

straight ϕ space equals space straight E space straight x spaceArea of the cross section of the gaussian cylinder.
Here, electric lines of force are parallel to the curved surface area of the cylinder, the flux due to the electric field of the plane sheet of charge passes only through two circular sections of the cylinder.


straight ϕ space equals space straight E space straight x space 2 straight A space space bold space bold space bold. bold. bold. bold space bold left parenthesis bold i bold right parenthesis

According to Gauss's Theorem,


straight ϕ space equals space straight q over straight epsilon subscript straight o

Here, charge enclosed by the gaussian surface,

space space space space space space space straight q space equals space straight sigma space straight A
therefore space space space straight ϕ space equals space σA over straight epsilon subscript straight o space space space space space bold space bold. bold. bold. bold space bold left parenthesis bold ii bold right parenthesis

From equations (i) and (ii), we get

straight E space straight x space 2 straight A space equals space σA over straight epsilon subscript straight o
space space space space space space space space space straight E space equals space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction

The direction of electric field for positive charge is in the outward direction and perpendicular to the plane of infinite sheet.

Direction of electric field for negative charge is in the inward direction and perpendicular to the sheet.

ii) Given: Two capacitors are in the ratio of 1:2.

That is, C2 = 2C1

When the capacitors are connected in parallel,

Total capacitance will be, CP = C1 + C2 = 3 C1

Energy stored in the capacitor,

straight E space equals space 1 half straight C subscript straight P space straight V subscript straight P squared
space space space space equals space fraction numerator 3 straight C subscript 1 space straight V subscript straight P squared over denominator 2 end fraction
When the capacitors are connected in series,

1 over straight C subscript straight S equals 1 over C subscript 1 plus 1 over C subscript 2

C subscript S space equals space fraction numerator 2 C subscript 1 over denominator 3 end fraction
Energy stored in the capacitor,

straight E space equals space 1 half straight C subscript straight S straight V subscript straight S squared

space space space equals space fraction numerator straight C subscript 1 straight V subscript straight S squared over denominator 3 end fraction

Given that, energy stored in both the cases is same.

That is,

fraction numerator 3 space straight C subscript 1 straight V subscript straight P squared over denominator 2 end fraction space equals space fraction numerator C subscript 1 V subscript S squared over denominator 3 end fraction
space space space space space space space space space V subscript P over V subscript s space equals space fraction numerator square root of 2 over denominator 3 end fraction

Hence, the result.
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(a) Deduce the expression for the torque acting on a dipole of dipole moment p in the presence of a uniform electric field E⃗.

(b) Consider two hollow concentric spheres, S1 and S2, enclosing charges 2Q and 4Q respectively as shown in the figure.

(i) Find out the ratio of the electric flux through them.

(ii) How will the electric flux through the sphere S1 change if a medium of dielectric constant 'εr' is introduced in the space inside S1 in place of air ?

 Deduce the necessary expression.  


a) Dipole in a uniform electric field:

Consider an electric dipole consisting of charges −q and +q and of length 2a placed in a uniform electric field making an angle θ with the electric field.

Forces acting on the two charges of the dipole, are +qE and -qE.

That is, the net force on the dipole is equal and opposite.

So, Force = 0

Two forces are equivalent to torque having magnitude given by,

 

Therefore, torque acting on the dipole is given by, 

b) i)  Charge enclosed by sphere S1 = 2Q

Charge enclosed by sphere S2 = 2Q + 4Q = 6Q

Now, using Gauss law, electric flux enclosed by sphere S1 and S2 is given by, 

 

ii) If a medium of dielectric constant 'εr' is introduced in the space inside S1 in place of air, electric flux becomes



 
That is, electric flux decreases.

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(i) If two similar large plates, each of area A having surface charge densities 1s and 2s are separated by a distance d in air, find the expressions for
(a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case.
(b) the potential difference between the plates.
(c) the capacitance of the capacitor so formed.
(ii) Two metallic spheres of radii R and 2R are charged so that both of these have same surface charge density s. If they are connected to each other with a conducting wire, in which direction will the charge flow and why ?


i)
  

Given, each plate has an area A and surface charge densities 1s and 2s respectively.

The plates 1 and 2 be separated by a small distance d.

For plate 1:

Surface charge density, straight sigma space equals space straight Q over straight A
For plate 2:

Surface charge density, straight sigma space equals space fraction numerator negative straight Q over denominator straight A end fraction equals negative straight sigma
Electric field in different regions:

Outside region 1,

E = fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction minus fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction equals 0

Outside region 2,

straight E equals space space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction minus fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction equals 0

Inner region:

In the inner region between plates 1 and 2, electric fields due to the two charged plates add up.

That is,

straight E space equals space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction plus space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction
space space space space equals straight sigma over straight epsilon subscript straight o
space space space space equals space fraction numerator straight Q over denominator straight epsilon subscript straight o straight A end fraction

Direction of electric field is from positive to the negative plate.

b) Potential difference between the plates, V = Ed = 1 over straight epsilon subscript straight o fraction numerator Q d over denominator A end fraction
c) Capacitance of the capacitor so formed, C = straight Q over straight V equals fraction numerator epsilon subscript o A over denominator d end fraction
ii) Potential on and inside of charged sphere is given by,

straight V space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction. straight q over straight r

space space space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction. fraction numerator 4 πr squared space straight sigma over denominator straight r end fraction

therefore space straight V space proportional to space straight r

That is, the sphere having larger radius will be at a higher potential.

Therefore, charge will flow from bigger sphere to smaller sphere.

Tips: -



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A long cylindrical shell carries positive surface charge  in the upper half and negative surface charge  in the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale)


D.

Field lines should originate from a positive charge and terminate negative charge. 

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Two charges, each equal to q, are kept at x = −a and x = a on the x-axis. A particle of mass m and charge qo =-q/2 is placed at the origin. If charge qo is given a small displacement (y<< a) along the y-axis, the net force acting on the particle is proportional to

  • y

  • -y

  • 1/y

  • 1/y


A.

y




Net force in negative y -direction,
Fnet = 2F cos θ 
 
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