Q26 of 26 Page 1

(a) Deduce the expression for the potential energy of a system of two charges q1 and q2 located at r and r respectively in an external electric field.

(b) Three points charges, + Q, + 2Q and – 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-points A1, B1 and C1 respectively, find the amount of the work done in shifting the charges to the new locations.


OR


Define electric flux. Write its S.I. unit. State and explain Gauss’s law. Find out the outward flux due to a point charge + q placed at the centre of a cube of side ‘a’. Why is it found to be independent of the size and shape of the surface enclosing it? Explain.


(a) Let q₁ and q₂ be the two charges located at r₁ and r₂, respectively, in an external electric field. The work done in bringing the chare q₁ from infinity to r₁ is W₁ = q₁V (r₁), where V(r₁) is the potential. Similarly, we can obtain the work done in bringing the charge q₁ from infinity to r₂.


Here, the work is done not only against the external field E but also against the field due to q₁. Hence, work done on q₂ against the external field is W2 = q2 V (r2),


Work done on q2 against the field due to q1, is given by



where r₁₂ is the distance between q₁ and q₂.


By the principle of superposition for fields, work done on q₂ against two fields will add with work done in bringing q₂ to r₂, which is given as


W2 + W12 = q2 V(r2 ) +


Thus, the potential energy of the system U = total work done in assembling the configuration


U = W₁ + W₂ + W₁₂


U = q₁V (r₁)+ q2 V(r2 ) +


(b) Given that the three charges are q1 = +Q, q2 = +2Q and q3 = -3Q. r = l (for each side)


Initial potential energy of system is given as



On substituting the value of all three charges



The charges are displaced to mid points then final potential energy of system,



Substituting the given values, we get



Work done, W = U2 – U1, we get



OR


Electric flux is defined as the total number of lines of force passing through the unit area of a surface perpendicularly.


Electric flux Δϕthrough an area element ΔS is given by


Δϕ= E. ΔS = EΔS cosθ


Where θ is the angle between E and ΔS. The SI unit of electric flux is N C-1m2.


According to the Gauss’s law the flux of the electric field through any closed surface S is 1/εₒ times the total charge enclosed by the surface.


Let the total flux through a sphere of radius r enclose a point charge q at its centre. Selecting a small area element as shown in the figure.



The flux through an area element ΔS is



The electric field due to a single charge q. The unit vector is along the radius vector from the centre of the sphere to the area element. Since, the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and have the same direction. Therefore,



As we know, the magnitude of the unit vector is 1, the total flux through the sphere is obtained by adding flux through all the area elements.



Since, each area element is at the same distance r from the charge, and the surface area element is the only varying quantity here, the summation will be only of all surface area element of the sphere. Now, S the total area of the sphere equals 4πr2 so, we can write flux as,



Hence, the above equation is an illustration of a general result of electrostatics called Gauss’s law.


Let a cube of side an enclose charge +q at its centre. Because the electric flux through the square surface is the square surfaces of cube are six. Hence, according to Gauss’s theorem in electrostatics, the total outward flux due to a charge +q of a cube is



The result shows that the electric flux passing through a closed surface is


proportional to the charge enclosed. In addition, the result reinforces that the flux is independent of the shape and size of the closed surface.


More from this chapter

All 26 →
22

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery.

A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time, the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected: (a) The electric field between the plates of the capacitor


(b) The energy stored in the capacitor


Justify your answer by writing the necessary expressions.


 23

Ajit had a high-tension tower erected on his farm land. He kept complaining to the authorities to remove it as it was occupying a large portion of his land. His uncle, who was a teacher, explained to him the need for erecting these towers for efficient transmission of power. As Ajit realized its significance, he stopped complaining. Answer the following questions:

(a) Why is it necessary to transport power at high voltage?


(b) A low power factor implies large power loss. Explain.


(c) Write two values each displayed by Ajit and his uncle.


 24

(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to ‘I’.

(B) Derive the expression for the magnetic field due to a solenoid of length ‘2 l’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distant ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’


OR


(a) Draw the magnetic field lines due to a circular loop of area Acarrying current I. Show that it acts as a bar magnet of magnetic momentum AI.


(b) Derive the expression for the magnetic field due to a solenoid of length ‘2 l’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distant ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’ ?


 25

(a) In Young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence deduce the expression for the fringe width.

(b) Show that the fringe pattern on the screen is actually a superposition of single slit diffraction from each slit.


(c) What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two slits is 1 mm ?


OR


(a) Two thin convex lenses L1 and L2 of focal lengths f1 and f2 respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens L1. Draw a ray diagram to show the image formation by the combination and hence derive the expression for the focal length of the combined system.


(b) A ray PQ incident on the face AB of a prism ABC, as shown in the figure, emerges from the face AC such that AQ = AR. Draw the ray diagram showing the passage of the ray through the prism. If the angle of the prism is 60°and refractive index of the material of the prism is 3, determine the values of angle of incidence and angle of deviation.