Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

NEET : Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

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Q. 35. Find the electric field strength vector if the potential of this field has the form φ = ar, where a is a constant vector, and r is the radius vector of a point of the field. 

Solution. 35. In accordance with the problem  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Thus from the equation Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 36. Determine the electric field strength vector if the potential of this field depends on x, y coordinates as
 (a) φ = a (x2  — y2);
 (b) φ = axy, 

where a is a constant. Draw the approximate shape of these fields .using lines of force (in the x, y plane).

Solution. 36. Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

So,  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

The sought shape of field lines is as shown in the figure (a) of answersheet assuming a > 0:

(b) Since φ = axy 

So,    Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Plot as shown in the figure (b) of answersheet


Q. 37. The potential of a certain electrostatic field has the form cp = a (x2 + y2) + bz2, where a and b are constants. Find the magnitude and direction of the electric field strength vector. What shape have the equipotential surfaces in the following cases: 

(a) a > 0, b> 0; (b) a > 0, b < 0? 

Solution. 37. Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

So,  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Hence  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Shape of the equipotential surface :

Put  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Then the equipotential surface has the equation  

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev and the equation of the equipotential surface is

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

which is an ellipse in p , z coordinates. In three dimensions the surface is an ellipsoid of revolution with semi- axis  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev then the equation is

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

This is a single cavity hyperboloid of revolution about z axis. If φ = 0 then 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

or   Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

is the equation of a right circular cone. 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

or  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

This is a two cavity hyperboloid of revolution about z-axis.


Q. 38. A charge q is uniformly distributed over the volume of a sphere of radius R. Assuming the permittivity to be equal to unity throughout, find the potential
 (a) at the centre of the sphere;
 (b) inside the sphere as a function of the distance r from its centre. 

Solution. 38. From Gauss’ theorem intensity at a point, inside the sphere at a distance r from the centre is given by,  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev and outside it, is given by Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

(a) Potential at the centre of the sphere,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

as  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

(b) Now, potential at any point, inside the sphere, at a distance r from it s centre.

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

On integration :  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 39. Demonstrate that the potential of the field generated by a dipole with the electric moment p (Fig. 3.4) may be represented as Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev where r is the radius vector. Using this expression, find the magnitude of the electric field strength vector as a function of r  and θ. 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 39. Let two charges +q and - q be separated by a distance l. Then electric potential at a point at distance r > > l from this dipole,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev     (1)

But  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

From Eqs. (1) and (2),

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

where p is magnitude of electric moment vector.

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 40. A point dipole with an electric moment p oriented in the positive direction of the z axis is located at the origin of coordinates. Find the projections Ez  and E of the electric field strength vector (on the plane perpendicular to the z axis at the point S (see Fig. 3.4)). At which points is E perpendicular to p? 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 40. From the results, obtained in the previous problem,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

From the given figure, it is clear that,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

and  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

When  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

So  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Thus Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev the points located on the lateral surface of the cone, having its axis, coinciding with the direction of z-axis and semi vertex angle Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 41. A point electric dipole with a moment p is placed in the external uniform electric field whose strength equals E0, with Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevIn this case one of the equipotential surfaces enclosing the dipole forms a sphere. Find the radius of this sphere.

Solution. 41. Let us assume that the dipole is at the centre of the one equipotential surface which is spherical (Fig.)* On an equipotential surface the net electric field strength along the tangent of it becomes zero. Thus

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Alternate : Potential at the point, near the dipole is given by,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

For φ to be constant,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Thus  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 42. Two thin parallel threads carry a uniform charge with linear densities λ and —λ. The distance between the threads is equal to l. Find the potential of the electric field and the magnitude of its strength vector at the distance r ≫ l at the angle θ to the vector l (Fig. 3.5).

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 42. Let P be a point, at distace r >> l and at an angle to θ the vector  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Also,  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 43. Two coaxial rings, each of radius R, made of thin wire are separated by a small distance l (l ≪ R) and carry the charges q and —q. Find the electric field potential and strength at the axis of the system as a function of the x coordinate (Fig. 3.6). Show in the same drawing the approximate plots of the functions obtained. Investigate these functions at |x| ≫ R.

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 43. The potential can be calculated by superposition. Choose the plane of the upper ring as x = l/2 and that of the lower ring as x = - l/2.

Then   Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

For  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

The electric field is  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

For  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev The plot is as given in the book.


Q. 44. Two infinite planes separated by a distance l carry a uniform surface charge of densities σ and —σ (Fig. 3.7). The planes have round coaxial holes of radius R, with 1 ≪ R. Taking the origin O and the x coordinate axis as shown in the figure, find the potential of the electric field and the projection of its strength vector E on the axes of the system as functions of the x coordinate. Draw the approximate plot φ (x).

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 44. The field of a pair of oppositely chaiged sheets with holes can by superposition be reduced to that of a pair of uniform opposite charged sheets and discs with opposite charges. Now the charged sheets do not contribute any field outside them. Thus using the result of the previous problem

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

The plot is as shown in the answersheet.


Q. 45. An electric capacitor consists of thin round parallel plates, each of radius R, separated by a distance l (l ≪ R) and uniformly charged with surface densities σ and —σ. Find the potential of the electric field and the magnitude of its strength vector at the axes of the capacitor as functions of a distance x from the plates if x ≫ l. Investigate the obtained expressions at x ≫ R. 

Solution. 45. For x > 0 we can use the result as given above and write

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

for the solution that vanishes at α. There is a discontinuity in potential for |x| = 0. The solution for negative x is obtained by σ → -σ. Thus

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Hence ignoring the jump

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

for large  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 46. A dipole with an electric moment p is located at a distance r from a long thread charged uniformly with a linear density λ. Find the force F acting on the dipole if the vector p is oriented
 (a) along the thread;
 (b) along the radius vector r;
 (c) at right angles to the thread and the radius vector r. 

Solution. 46. Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev not change as the point of observation is moved along the thread.

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 47. Find the interaction force between two water molecules separated by a distance l = 10 nm if their electric moments are oriented along the same straight line. The moment of each molecule equals p = 0.62.10-29  C • m. 

Solution. 47. Force on a dipole of moment p is given by,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

In our problem, field, due to a dipole at a distance l, where a dipole is placed,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Hence, the force of interaction,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 48. Find the potential φ (x, y) of an electrostatic field E = a (yi xj), where a is a constant, i and j are the unit vectors of the x and y axes. 

Solution. 48. 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

On integrating,  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 49. Find the potential φ (x, y) of an electrostatic field E = 2axyi + a (x2  — y2) j, where a is a constant, i and j are the unit vectors of the x and y axes. 

Solution. 49.  

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

or   Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

On integrating, we get,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 50. Determine the potential φ (x, y, z) of an electrostatic field E = ayi (ax + bz) j + byk, where a and b are constants, i, j, k are the unit vectors of the axes x, y, z. 

Solution. 50. Given, again

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

On integrating,

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 51. The field potential in a certain region of space depends only on the x coordinate as φ = — ax3 + b, where a and b are constants. Find the distribution of the space charge p (x).

Solution. 51. Field intensity along x-axis.

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev     (1)

Then using Gauss’s theorem in differential from 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 52. A uniformly distributed space charge fills up the space between two large parallel plates separated by a distance d. The potential difference between the plates is equal to Δφ. At what value of charge density p is the field strength in the vicinity of one of the plates equal to zero? What will then be the field strength near the other plate? 

Solution. 52. In the space between the plates we have the Poisson equation

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

or,  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

where p0 is the constant space charge density between the plates. 

We can choose  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Then  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRevConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

NowConstant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

if    Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

then Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

Also Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 53. The field potential inside a charged ball depends only on the distance from its centre as φ = ar2 + b, where a and b are constants. Find the space charge distribution p (r) inside the ball.

Solution. 53. Field intensity is along radial line and is

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev                      (1)

From the Gauss’ theorem, 

Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

where dq is the charge contained between the sphere of radii r and r + dr. 

Hence  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev  (2)

Differentiating  Constant Electric Field In Vacuum (Part - 3) - Electrodynamics, Irodov JEE Notes | EduRev

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