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Electric Charges and Fields

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Physics Part I

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Physics

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The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ = A/r , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is:

  • fraction numerator straight Q over denominator 2 πa squared end fraction
  • fraction numerator straight Q over denominator 2 straight pi left parenthesis straight b to the power of 2 minus end exponent straight a squared right parenthesis end fraction
  • fraction numerator 2 straight Q over denominator straight pi left parenthesis straight a squared minus straight b squared right parenthesis end fraction
  • fraction numerator 2 straight Q over denominator πa squared end fraction

A.

fraction numerator straight Q over denominator 2 πa squared end fraction

A Gaussian surface at distance r from centre.




fraction numerator straight Q space plus integral subscript straight a superscript straight r begin display style straight A over straight r end style 4 straight pi squared dr over denominator straight epsilon subscript straight o end fraction space equals space straight E 4 πr squared
straight E 4 straight epsilon subscript straight o straight r squared space equals space straight Q space plus space straight A fraction numerator 4 straight pi over denominator straight r squared end fraction open parentheses fraction numerator straight r squared minus straight a squared over denominator 2 end fraction close parentheses
straight E space equals fraction numerator 1 over denominator 4 πε subscript straight o end fraction space open square brackets straight Q over straight r squared plus straight A space 2 straight pi open parentheses fraction numerator straight r squared minus straight a squared over denominator straight r squared end fraction close parentheses close square brackets
straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction open square brackets straight Q over straight r squared plus straight A 2 straight pi minus fraction numerator straight A 2 πa squared over denominator straight r squared end fraction close square brackets
straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight x space straight A space straight x space 2 straight pi
At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is 
As, Q = 2πAa2
i.e A = Q/2πa2

A Gaussian surface at distance r from centre.




fraction numerator straight Q space plus integral subscript straight a superscript straight r begin display style straight A over straight r end style 4 straight pi squared dr over denominator straight epsilon subscript straight o end fraction space equals space straight E 4 πr squared
straight E 4 straight epsilon subscript straight o straight r squared space equals space straight Q space plus space straight A fraction numerator 4 straight pi over denominator straight r squared end fraction open parentheses fraction numerator straight r squared minus straight a squared over denominator 2 end fraction close parentheses
straight E space equals fraction numerator 1 over denominator 4 πε subscript straight o end fraction space open square brackets straight Q over straight r squared plus straight A space 2 straight pi open parentheses fraction numerator straight r squared minus straight a squared over denominator straight r squared end fraction close parentheses close square brackets
straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction open square brackets straight Q over straight r squared plus straight A 2 straight pi minus fraction numerator straight A 2 πa squared over denominator straight r squared end fraction close square brackets
straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight x space straight A space straight x space 2 straight pi
At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is 
As, Q = 2πAa2
i.e A = Q/2πa2

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 (a) An electric dipole of dipole moment bold italic p with rightwards harpoon with barb upwards on top consists of point charges +q and –q separated by a distance 2a apart. Deduce the expression for the electric field bold italic E with rightwards harpoon with barb upwards on top due to the dipole at a distance x from the center of the dipole on its axial line in terms of the dipole moment  bold italic p with rightwards harpoon with barb upwards on top. Hence show that in the limit x >> a, bold italic E with rightwards harpoon with barb upwards on top space rightwards arrow  2 straight p with rightwards harpoon with barb upwards on top/ (4p straight epsilon0 x3).

(b) Given the electric field in the region  bold italic E with rightwards harpoon with barb upwards on top= 2x i , find the net electric flux through the cube and the charge enclosed by it.


Electric field on axial line of an electric dipole is given by, 

Suppose, P is a point at distance r from the center of the dipole on the side of charge –q.

Electric field at P due to –q is given by, 

E subscript negative q end subscript space equals space minus fraction numerator q over denominator 4 pi epsilon subscript o left parenthesis r minus a right parenthesis squared end fraction p with hat on top space semicolon space bold italic p with hat on top is the unit vector along the dipole axis. 

Electric field at P due to +q is given by,

straight E subscript plus straight q end subscript straight space equals straight space fraction numerator straight q over denominator 4 πε subscript straight o left parenthesis straight r minus straight a right parenthesis squared end fraction straight space straight p with hat on top 

Therefore, total electric field at point P is,

Error converting from MathML to accessible text.
b) Since, the electric field is parallel to the faces parallel to xy and xz planes, the electric flux through them is zero.

Electric flux through the left face, ϕ subscript L space equals space left parenthesis E subscript L right parenthesis left parenthesis a squared right parenthesis space cos 180 to the power of 0 space equals 0
Electric flux through the right face, 

ϕ subscript R space equals space left parenthesis E subscript R right parenthesis left parenthesis a squared right parenthesis space cos 0 to the power of 0 space equals left parenthesis 2 a right parenthesis space left parenthesis a squared right parenthesis cross times 1 space equals space 2 a cubed

Net flux is given by, straight ϕ space equals space space q subscript e n c l o s e d end subscript over epsilon subscript o space equals space 2 straight a cubed

rightwards double arrow                               straight space straight q subscript enclosed straight space equals straight space 2 straight a cubed straight epsilon subscript straight o

3782 Views

(a) Define electric flux. Write its S.I. units.

(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

(c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?     

a) Electric flux is defined as the total number of electric field lines passing through an area normal to them.

straight phi straight space equals straight space contour integral straight E with rightwards harpoon with barb upwards on top. ds with rightwards harpoon with barb upwards on top
SI unit is Nm2/C.

 b) Gauss’s theorem states that the total electric flux through a closed surface is equal to  times the net charge enclosed by the surface. 

 

Consider a uniformly charged infinite plane sheet of charge density straight sigma .

Consider a Gaussian surface inside as shown in the figure which is in the form of a cylinder.

On applying Gauss’s law, we have

straight phi equals contour integral straight E with rightwards harpoon with barb upwards on top. ds with rightwards harpoon with barb upwards on top straight space equals straight space σds over straight epsilon subscript straight o

rightwards double arrow straight space straight E. ds straight space plus straight space straight E. ds straight space plus straight space 0

rightwards double arrow straight space fraction numerator straight sigma. ds over denominator straight epsilon subscript straight o end fraction equals 2 straight E. ds straight space equals straight space fraction numerator straight sigma. ds over denominator straight epsilon subscript straight o end fraction
rightwards double arrow straight space straight E straight space equals straight space fraction numerator straight sigma over denominator 2 straight epsilon subscript straight o end fraction 

Thus, electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge.

If the surface charge density s is negative the electric field is directed towards the surface charge.

c) i) Away from the charged sheet.

ii) towards the plane sheet.

3453 Views

Using Gauss’ laws deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point (i) outside and (ii) inside the shell.

Plot a graph showing variation of electric field as a function of r > R and r < R. (r being the distance from the centre of the shell). 


i) Consider a uniformly charged thin spherical shell of radius R carrying charge Q. To find the electric field outside the shell, we consider a spherical Gaussian surface of radius (r >R), concentric with given shell. If E is electric field outside the shell, then by symmetry electric field strength has same magnitude E0 on the Gaussian surface and is directed radially outward.

So, electric flux through Gaussian surface is given by, 

contour integral subscript s space equals space stack E subscript o with rightwards harpoon with barb upwards on top space. space d S with rightwards harpoon with barb upwards on top space

Therefore, 

contour integral space equals space E subscript o space d s space c o s space 0 space equals space E subscript o.4 pi r squared space 

Charge enclosed by the Gaussian surface is Q.

Therefore, using gauss’s theorem, we have

contour integral subscript s space equals space E with rightwards harpoon with barb upwards on top subscript o space d S space equals space 1 over epsilon subscript o space x space c h a r g e space e n c l o s e d

rightwards double arrow space E subscript o space 4 pi r squared space equals space 1 over epsilon subscript o x space Q

rightwards double arrow space space E subscript o italic space italic equals italic space fraction numerator italic 1 over denominator italic 4 pi epsilon subscript o end fraction Q over r to the power of italic 2 

Thus, electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre.

ii) Electric field inside the shell:

 

The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical shell. Let’s consider a spherical Gaussian surface of radius (r < R). If E is the electric field inside the shell, then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially outward. 

Electric flux through the Gaussian surface is given by, 

equals space integral subscript S space E with rightwards harpoon with barb upwards on top subscript i. space d S with rightwards harpoon with barb upwards on top
 
integral E subscript i space d S space c o s space 0 space equals space E subscript i space. space 4 pi r squared
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.

Therefore, using Gauss’s theorem, we have

space space space space space space integral subscript straight S space E with rightwards harpoon with barb upwards on top subscript i. space d S with rightwards harpoon with barb upwards on top space equals space 1 over epsilon subscript o x space c h a r g e space e n c l o s e d

rightwards double arrow space E subscript i.4 pi r squared space equals space 1 over epsilon subscript o space x space 0
rightwards double arrow space E subscript i space equals space 0 space 

Thus, electric field at each point inside a charged thin spherical shell is zero.

 

The graph above shows the variation of electric field as a function of R.

2736 Views

While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy at his residence. The boy insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child.
Answer the following questions based on the above information:

(a) Why is it safer to sit inside a car during a thunderstorm?

(b) Which two values are displayed by Dr. Pathak in his actions?

(c) Which values are reflected in parents’ response to Dr. Pathak?

(d) Give an example of a similar action on your part in the past from everyday life.           


a) On the basis of electrostatic screening, no electric field exists inside the charged conducting body. During lightening a shower of the charged particles falls on the earth. So it would be safer to sit inside the car.

(b) Dr. Pathak knows the result of lightening during thunderstorm; so he displayed two actions;

(i) Shows love, kindness and sympathy to the child.

(ii) Keeping in view the safety of the child, he allow the boy to sit in the car till the thunderstorm stopped.

(c) Parent meets Dr. Pathak; and express their gratitude and heart felt thank for providing the safety to the child from lightning and thunderstorm.

(d) Many of us have read in the newspaper that the person either working in the field or in open space have lost their life during thunderstorm. So the persons belonging to villages must be given advices that they should remain inside the houses during thunderstorm.

1518 Views