Video Lecture

Theory For Notes Making

Electric Lines of Force

The lines of forces are purely geometrical construction which helps in visualising the nature of electric field in a region. It has no physical existence. Lines of forces are drawn in such a way that the tangent to a line of force gives the direction of resultant electric field. The density of field lines in any region is proportional to the magnitude of the electric field in that region. Field lines originate on positive charges and terminate on negative charges.

1.

There cannot be any closed line of force in an electrostatic field \displaystyle \vec{E}. Electric lines of force emerge from positive charges and terminate on negative charges (or extends to infinity).

2.     

Crowded lines represent strong field, while distant represent weak field.

3.

The number of lines originating or terminating on a charge is proportional to the magnitude of the charge. In SI units, the number of electric field lines associated with a unit charge (i.e., 1 coulomb) is taken as 1/ε0. Thus, if a body encloses a charge q, total lines of force associated with it (also called flux) will be q0.

In the above figure the magnitude of q2 is more than that of q1.

4.

Lines of force can never cross each other. It is obvious, because if they cross at a point the intensity at that point will have two directions, which is absurd.

5.

Lines of force have tendency to contract longitudinally (like a stretched elastic string) and repel each other laterally. This concept, like the lines of force, is also imaginary. It helps in understanding how attraction is produced between opposite charges, and how repulsion is produced between similar charges.

Electric field lines for two positive point-charges of equal magnitude.

Electric field lines of two similar charges placed near to each other

6.  

If there is no electric field in a region of space, there will be no lines of force. This is why inside a conductor or at a neutral point (where resultant intensity is zero) there cannot be any line of force.

7.

If the lines of force are equidistant straight lines, the field is uniform [Fig. (A)]. If either lines of force are not equidistant or straight lines or both, the field will be non-uniform [Fig. (B)]

Electric Flux

Consider a surface of area A placed in an electric field E.

Now the number of lines that penetrate through the surface depends on three factors

(i) Strength of electric field E

(ii) Surface area A

(iii) Orientation of area with respect to the electric field .

If N is the number of electric field lines passing through the Area then

\displaystyle N\propto \vec{E} —————(i)

And \displaystyle N\propto \vec{A}   —————(ii)

On combining (i) and (ii) we get

\displaystyle N\propto \vec{E}.\vec{A}

Here the term \displaystyle \vec{E}.\vec{A}  is called electric flux and is denoted by \displaystyle \phi . So we can define electric flux as the dot product of electric field vector and area vector. The number of electric field lines passing through the surface area depends on the electric flux.

Hence \displaystyle \phi =\vec{E}.\vec{A}

Or \displaystyle \phi =EA\cos \theta .

Here   \displaystyle A\cos \theta  is the projection of area A onto a plane oriented perpendicular to the electric field as shown in the diagram given below and q is the angle between electric field vector and normal to the area.

From the SI units of E and A, we see that \displaystyle \phi  has units of newton meters squared per coulomb (N.m2/C). Also remember that \displaystyle \phi  is a scalar quantity.      

From the equation \displaystyle \phi =EA\cos \theta  we can conclude that

(i)

when θ = 0 that is the normal to area is parallel to the electric field or the plane of the area is perpendicular to the field , the flux through the area is maximum

(ii)

When θ =90o the flux is zero.

In case of closed surface the flux may be positive or negative depending on the direction of field lines and the angle between the normal and the electric field as shown in the diagram given below. In the diagram there are two close surfaces. The first one contains a positive charge and the second contains a negative charge. In the first case the angle between the normal to surface and the electric field is less than 90o. So according to formula \displaystyle \phi =EA\cos \theta , the flux is positive. whereas in the second case the flux is negative because θ is more than 90o.

Important Points

(i)  

For a closed body outward flux is taken to be positive [Fig. (A)] while inward negative
[Fig. (B)]

(ii)       

The electric flux is contributed by only that charge which is enclosed by the surface. If a charge is lying outside the closed surface the net electric flux passing the closed surface is zero whereas electric field is produced by all the charges whether inside on outside.

Objective Assignment

Q.1

The electric field in a region of space is given by =  N/C. The flux of  due to this field through an area 2 m2 lying in the y-z plane, in SI units, is

(a)  10        

(b)    20       

(c)  10 \sqrt{2}     

(d)  2 \sqrt{{29}}

Ans. (a)

Q.2

Figure shows a closed surface which intersects a conducting sphere. If a positive charged is placed at the point P, the flux of the electric field through the closed surface

(a) will remain zero                

(b) will become positive

(c) will become negative        

 (d) will become undefined

Ans. (b)

Q.3       

In a region of space, the electric field is in the x-derection and proportional to s, i.e., =E0xî. consider an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge inside this volume is

(a) zero                                   

(b) e0E0a3                                                                                           

(c) \frac{1}{{\varepsilon 0}} E0a3                           

(d) \frac{1}{6}e0E0a2

Ans. (b)

Q.4

In a region of space, the electric field is given by . The electric flux through a surface of area of 100 units in x-y plane is 

(a)    800 units           

(b)     300 units             

(c)    400 units          

(d)    1500 units

Ans. (b)

Q.5

A cylinder of length L and radius b has its axis coincident with the x The electric field in this region is E = 200j. Find the flux through  thebottom of the cylinder.

(a)– 200 pb2    

(b)– 700 pb2                                

(c)-400bL       

(d) –100 pb2

Ans: (c)

Q.6

Two oppositely charged parallel plates have dimensions of 10x20cm and have a charge of 20mC each. Find the electric flux and electric flux density between the two plates.

(a)0.25×10Nm2/C  , 0.125×108 N/C

(b) 25×10Nm2/C  , 125×108 N/C

(c) 2.25×10Nm2/C  , 1.125×108 N/C     

(d)35×10Nm2/C  , 2×108 N/C

Ans: (c)

7.

A long string with a charge of  \lambda per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be

(a) \lambda a/{{\varepsilon }_{0}}           

(b) \frac{{\sqrt{2}\,\lambda a}}{{{{\varepsilon }_{0}}}}                  

(c) \frac{{6\lambda {{a}^{2}}}}{{{{\varepsilon }_{0}}}}                     

(d) \frac{{\sqrt{3}\,\lambda a}}{{{{\varepsilon }_{0}}}}

Ans (d)

Subjective Assignment

Q.1     

A hemisphere of radius 20cm is kept in a uniform electric field of 200N/C. If its circular base is perpendicular to the electric field find the electric flux passing through the hemisphere.

Q.2     

A charge of 10mC is kept at the center of a hemisphere of radius 20cm.Find the electric flux passing through it.

Q.3     

A charge of 20mC is kept at a distance of 10cm above the center of a square plane of side 20cm lying horizontally. Find the electric flux passing through the plane.

Q.4     

A charge of 6mC is kept at a distance of 20cm from the center of a circular disc of radius 5cm.Find the electric flux passing through the disc.

Q.5     

A line of charge of infinite length and linear charge density 10mC/m is kept coaxial with a cylinder of radius 5cm and length 50cm. Find the electric flux passing through the cylinder.

Q.6     

Three charges of 10mC , -40mC and 50mC are enclosed by a surface. Find the net flux passing through the surface.

Q..7    

Two oppositely charged parallel plates have dimensions of 10x20cm and have a charge of 20mC each. Find the electric flux and electric flux density between the two plates.

Q.8     

The electric field in a    region    is    given    by E = 3/4Eo i + 4/3 Eo j with   Eo= 6000N/C. Find the flux passing through   a rectangular surface of area 4m2 parallel to the x-z plane

Q.9     

A triangular wedge shaped body is kept as shown. If an electric field given by E= 20i + 10j + 5k is existing in that region and all the dimensions are in meters then find the electric flux passing through all of its faces.

Q.10   

The electric field in a region is given by = E0 +E0 with E0 = 2.0´103 N/C. find the flux of this field through a rectangular surface of area 1.2 m2 parallel to the Y-Z plane.

Q.11   

A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube.

Q.12   

Show that there can be no net charge in a region in which the electric field is uniform at all points.

Q.13   

The electric field in a region is given by = . Find the charge contained inside a cubical volume bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5´103 N/C, l=2cm and a = 1 cm.

Q.14   

A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.

Q.15   

A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in figure. Find the flux of the electric field through the square surface.

Q.16   

Find the flux of the electric field through a spherical surface of radius R due to a charge of 107 C at the centre and another equal charge at point 2R away form the centre.

Q.17   

A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and the Gauss’s law, find the flux of the electric field due to this charge through the surface of the hemisphere.

Q.18   

A spherical volume contains a uniformly distributed chargeof density 2.0´104 C/m3. Find the electric field at a point inside the volume at a distance 4.0 cm v   from the centre.

Q.19         

Write an expression for the flux \displaystyle \Delta \phi , of the electric field \displaystyle \vec{E}, through an area element \displaystyle \vec{E}.

Q.20         

Two charges of magnitudes–2Q and +Q are located at points (a, 0) and (4a, 0) respectively.  What is the electric flux due to these charges through a sphere of radius ‘3a’ and ‘5a’ respectively with its centre at the origin?

Q.21         

Define electric flux.  Write its S.I.unit.  A charge q is enclosed by a spherical surface of radius R.  If the radius is reduced to half, how would the electric flux through the surface change ?

Q.22   

Given a uniform electric field \displaystyle \vec{E}=4\times \,{{10}^{3}}\,\hat{i}\,N/C.  Find the flux of this field through a square of 5 cm on a side whose plane is parallel to the Y-Z plane.  What would be the flux through the same square if the plane makes a 30º angle with the x-axis ?

Q.23         

Consider a uniform electric field E = 3 × 103 î N/C.

(a)      What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?

(b)      What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

(c)      What is the net flux of the uniform electric field through a cube of side 20cm oriented so that its faces are parallel to the coordinate planes?