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# Doc: Electric Field and Electric Field Lines Class 12 Notes | EduRev

## JEE : Doc: Electric Field and Electric Field Lines Class 12 Notes | EduRev

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ELECTRIC FIELD
The figure showed a charge q is lying in free space.

Now a charge q C is brought near it.
By coulomb's law we know that the charge q experiences a force and it exerts an equal force on q C.
How does q become aware of the presence of q C?
(We don't expect q to have sensory organs just as we have)
The answer is electric field !!!
Electric field is the space surrounding an electric charge q in which another charge qC experiences an (electrostatic) force of attraction, or repulsion.

The direction of electric field is radially outwards for a positive charge and is radially inwards for a negative charge as shown in the figure above. There are some points to always to be kept in mind. These are

1. The electric field can be defined as a space surrounding a charge in which another static charge experiences a force on it.

2. In a region electric field is said to exist if an electric force is exerted on a static charge placed at that point.

3. It is important to note that with every charged particle, there is an electric field associated which extends up to infinity.

4. No charged particle experiences force due to its own electric field.

=

A very small positive charge which does not produce its significant electric field is called a test charge.

Thus electric field strength at point can be defined generally as "Electric field strength at only point in space to be the electrostatic force per unit charge on a test charge."

If a charge q0 placed at a point in an electric field, experiences a net force on it, then electric field strength at that point can be

or  ....(1) [q0C test charge]

(a) Electric Field Strength due to Point Charge:

As discussed earlier, if we find the electric field due to a point charge at a distance x from it. Its magnitude can be given as

...(2)

(b) Vector Form of Electric field due to a Point Charge:

As shown in figure, the direction of electric field strength

at point P is along the direction of . Thus the value of

can be written as

or  ...(3)

It should be noted that the expression in equation (2) and (3) are only valid for point charges. We can not find electric field strength due to charged extended bodies by concentrating their whole charge at geometric centre and using the result of a point charge.

GRAPH OF ELECTRIC FIELD DUE TO BINARY CHARGE CONFIGURATION

1.
2.

3.
4.

ELECTRIC FIELD STRENGTH AT A GENERAL POINT DUE TO A UNIFORMLY CHARGED ROD
As shown in figure, if P is any general point in the surrounding of rod, to find the electric field strength at P, again we consider an element on rod of length dx at a distance x from point O as shown in figure.

Now if dE be the electric field at P due to the element, then it can be given as

Here dq =

Now we resolve electric field in components.

Electric field strength in x-direction due to dq at P is dEx = dE sin q

or

=

Here we have x = r tan q

and dx = r sec2 qdq

Thus we have

Strength =

Net electric field strength due to dq at point P in x-direction is

or

or

Similarly, the electric field strength at point P due to dq in y-direction is

dEy = dE cos q

or dEy =

Again we have x = r tan q

and dx = r sec2 q dq

Thus we have dEy =  =

Net electric field strength at P due to dq in y-direction is

Ey =

or Ey =

or Ey =

Thus electric field at a general point in the surrounding of a uniformly charged rod which subtend angles q1 and q2 at the two corners of the rod can be given as

in ||-direction Ex =  =

in ^ -direction Ey =  =
1. r is the perpendicular distance of the point from the wire
2. q1 and q2 should be taken in opposite sense

Electric Field Due To Infinite Wire (L >> R):
Here we have to find the electric field at point p due to the given infinite wire. Using the formula learnt in above section which

For above case,

Enet at

Electric field due to semi infinite wire:

For this case

Enet at s

ELECTRIC FIELD DUE TO A UNIFORMLY CHARGED RING
Case - I: At its Centre
Here by symmetry we can say that electric field strength at centre due to every small segment on ring is cancelled by the electric field at centre due to the segment exactly opposite to it. As shown in figure. The electric field strength at centre due to segment AB is cancelled by that due to segment CD. This net electric field strength at the centre of a uniformly charged ring is zero

Case II: At a Point on the Axis of Ring
For this look at the figure. There we'll find the electric field strength at point P due to the ring which is situated at a distance x from the ring centre. For this we consider a small section of length dl on ring as shown. The charge on this elemental section is

dq =  dl [Q = total charge of ring]

Due to the element dq, electric field strength dE at point P can be given as

The component of this field strength dE sin which is normal to the axis of the ring will be cancelled out due to the ring section opposite to dl. The component of the

electric field strength along the the axis of ring dE cos due to all the sections will be added up. Hence total electric field strength at point P due to the ring is

=

or

=

=

EP =

ELECTRIC FIELD STRENGTH DUE TO A UNIFORMLY SURFACE CHARGED DISC
If there is a disc of radius R, charged on its surface with surface charge density s C/m2, we wish to find electric field strength due to this disc at a distance x from the centre of disc on its axis at point P shown in figure.

To find electric field at point P due to this disc, we consider an elemental ring of radius y and width dy in the disc as shown in figure. Now the charge on this elemental ring dq can be given as

dq = s 2p y dy [Area of elemental ring ds = 2py dy]

Now we know that electric field strength due to a ring of radius R. Charge Q at a distance x from its centre on its axis can be given as

[As done earlier]

Here due to the elemental ring electric field strength dE at point P can be given as

=

Net electric field at point P due to this disc is given by integrating above expression from O to R as

= Kspx  = 2Kspx

E =

Case : (i) If x >> R

=  =  =

i.e. behaviour of the disc is like a point charge.

Case :(ii) If x << R

i.e. behaviour of the disc is like infinite sheet.

Electric Field Strength due to a Uniformly charged Hollow Hemispherical Cup:
Figure shows a hollow hemisphere, uniformly charged with surface charge density  sC/m2. To find electric field strength at its centre C, we consider an elemental ring on its surface of angular width dq at an angle q from its axis as shown. The surface area of this ring will be

ds = 2pR sin q × Rdq

Charge on this elemental ring is

dq = sds = s. 2pR2 sin q dq

Now due to this ring electric field strength

at centre C can be given as

=

= pKs sin 2q dq

Net electric field at centre can be obtained by integrating this expression between limits 0 to as

E0 =  =  =

CONSERVATIVE FORCE
A force is said to be conservative if work done by or against the force in moving a body depends only on the initial and final positions of the body and not on the nature of path followed between the initial and final positions.

Consider a body of mass m being raised to a height h vertically upwards as shown in above figure. The work done is mgh. Suppose we take the body along the path as in (b). The work done during horizontal motion is zero. Adding up the works done in the two vertical path of the paths, we get the result mgh once again. Any arbitrary path like the one shown in (c) can be broken into elementary horizontal and vertical portions. Work done along the horizontal path is zero. The work done along the vertical parts add up to mgh. Thus we conclude that the work done in raising a body against gravity is independent of the path taken. It only depends upon the initial and final positions of the body. We conclude from this discussion that the force of gravity is a conservative force.

Examples of Conservative forces.

(i) Gravitational force, not only due to Earth due in its general form as given by the universal law of gravitation, is a conservative force.

(ii) Elastic force in a stretched or compressed spring is a conservative force.

(iii) The electrostatic force between two electric charges is a conservative force.

(iv) Magnetic force between two magnetic poles is a conservative force.

Forces acting along the line joining the centres of two bodies are called central forces. Gravitational force and Electrostatic forces are two important examples of central forces. Central forces are conservative forces.

Properties of Conservative forces:

Work done by or against a conservative force depends only on the initial and final positions of the body.

Work done by or against a conservative force does not depend upon the nature of the path between initial and final positions of the body.

If the work done by a force in moving a body from an initial location to a final location is independent of the path taken between the two points, then the force is conservative.

Work done by or against a conservative force in a round trip is zero.

If a body moves under the action of a force that does not totally work during any round trip, then the force is conservative; otherwise, it is non-conservative.

The concept of potential energy exists only in the case of conservative forces.

The work done by a conservative force is completely recoverable.

Complete recoverability is an important aspect of the work of a conservative force.

Work done by conservative forces:

Ist format : (When constant force is given)

Ex.33 Calculate the work done to displace the particle from (1, 2) to (4, 5). if

Sol. dw = ( )

dw =  ⇒ dw = 4dx 3dy

=     ⇒ w =

w = (16 - 4) (15 - 6) ⇒ w = 12 9 = 21 Joule

II format : (When F is given as a function of x, y, z)

If

then

dw =  ⇒ dw = Fxdx Fydy FZdz

Ex.34 An object is displaced from position vector  to  under a force . Find the work done by this force.

Sol. =  Ans.

IIIrd format (perfect differential format)

Ex.35 If  then find out the work done in moving the particle from position (2, 3) to (5, 6)

Sol. dw =

dw =

dw = ydx xdy

Now ydx xdy = d(xy) (perfect differential equation)

⇒ dw = d(xy)

for total work done we integrate both side

Put xy = k

then at (2, 3) ki = 2 × 3 = 6

at (5, 6) kf = 5 × 6 = 30

then w =  ⇒ w = (30 - 6) = 24 Joule

NON-CONSERVATIVE FORCES

A force is said to be non-conservative if work done by or against the force in moving a body depends upon the path between the initial and final positions.

The frictional forces are non-conservative forces. This is because the work done against friction depends on the length of the path along which a body is moved. It does not depend only on the initial and final positions. Note that the work done by frictional force in a round trip is not zero.

The velocity-dependent forces such as air resistance, viscous force, magnetic force etc., are nonconservative forces.

Difference between conservative and Non-conservative forces

 S.No. Conservative forces Non-Conservative forces 1 Work done does not depend upon path Work done depends on path. 2 Work done in round trip is zero. Work done in a round trip is not zero. 3 Central in nature. Forces are velocity- dependent and retarding in nature. 4 When only a conservative force is applied to a system, the kinetic energy and potential energy can change. However their sum, the mechanical energy of the system, does not change, Work done against a non- conservative force may be dissipated as heat energy. 5 Work done is completely recoverable. Work done is not completely recoverable.

ELECTROSTATIC POTENTIAL ENERGY:
(a) Electrostatic Potential Energy:

Potential energy of a system of particles is defined only in conservative fields. As the electric field is also conservative, we define potential energy in it. Before proceeding further, we should keep in mind the following points, which are useful in understanding potential energy in electric fields.

(i) Doing work implies supply of energy

(ii) Energy can neither be transferred nor be transformed into any other form without doing work

(iii) Kinetic energy implies utilization of energy whereas potential energy implies storage of energy

(iv) Whenever work is done on a system of bodies, the supplied energy to the system is either used in form of KE of its particles or it will be stored in the system in some form, increases the potential energy of system.

(v) When all particles of a system are separated far apart by infinite distance there will be no interaction between them. This state we take as reference of zero potential energy.

Now potential energy of a system of particles we define as the work done in assembling the system in a given configuration against the interaction forces of particles.

Electrostatic potential energy is defined in two ways.

(i) Interaction energy of charged particles of a system.

(ii) Self-energy of a charged object (will be discussed later)

(b) Electrostatic Interaction Energy :

Electrostatic interaction energy of a system of charged particles is defined as the external work required to assemble the particles from infinity to a given configuration. When some charged particles are at infinite separation, their potential energy is taken zero as no interaction is there between them. When these charges are brought close to a given configuration, external work is required if the force between these particles is repulsive and energy is supplied to the system hence final potential energy of system will be positive. If the force between the particles is attractive work will be done by the system and final potential energy of system will be negative.

Let us take some illustrations to understand this concept in detail.

(c) Interaction Energy of a System of Two Charged Particles :

[-ve sign shows that x is decreasing]

[Interaction energy]

Motion of a Charged Particle and Angular Momentum Conservation:

We know that a system of particles when no external torque acts, the total angular momentum of system remains conserved. Consider following examples which explains the concept for moving charged particles.

Ex.40 Figure shows a charge Q fixed at a position in space. From a large distance another charged particle of charge q and mass m is thrown toward Q with an impact parameter d as shown with speed v. find the distance of closest approach of the two particles.

Sol.

Here we can see that as q moves toward Q, a repulsive force acts on -q radially outward Q. Here is the line of action of force passes through the fixed charge, no torque act on q relative to the fixed point charge Q, thus here we can say that with respect to Q, the angular momentum of q must remain constant. Here we can say that q will be closest to Q when it is moving perpendicular to the line joining the two charges as shown.

If the closest separation in the two charges is rmin, from conservation of angular momentum we can write

mvd = mv0 rmin ...(1)

Now from energy conservation, we have

Here we use from equation (1) v0 =

(2)

Solving equation (2) we'll get the value of rmin.

Potential Energy for a System of Charged Particles:

When more than two charged particles are there in a system, the interaction energy can be given by sum of interaction energy of all the pairs of particles. For example, if a system of three particles having charges q1, q2 and q3 is given as shown in figure. The total interaction energy of this system can be given as

Derivation for a system of point charges:

(i) Keep all the charges at infinity. Now bring the charges one by one to its corresponding position

and find work required. PE of the system is algebraic sum of all the works.
Let W1 = work done in bringing first charge

W2 = work done in bringing the second charge against force due to 1st charge

W3 = work done in bringing third charge against force due to 1st and 2nd charge.

PE = W1 +  W2 +  W3  .................. (This will contain  = nC2 terms)

(ii) Method of calculation (to be used in problems)

U = sum of the interaction energies of the charges.

= (U12 +  U13  ........ U1n) (U23 +  U24  .............. U2n) (U34 +  U35  .........U3n) ........

(iii) Method of calculation useful for symmetrical point charge systems.

Find PE of each charge due to rest of the charges.

If U1 = PE of first charge due to all other charges.

= (U12 +  U13  ......... U1n)

U2 = PE of second charges due to all other charges.

= (U21 +  U23  .......... U2n)

U = PE of the system =

ELECTRIC POTENTIAL
Electric potential is a scalar property of every point in the region of electric field. At a point in an electric field, electric potential is defined as the interaction energy of a unit positive charge.

If at a point in electric field a charge q0 has potential energy U, then electric potential at that point can be given as

V = U/qo   joule/coulomb

As potential energy of a charge in an electric field is defined as work done in bringing the charge from infinity to the given point in electric field. Similarly, we can define electric potential as "work done in bringing a unit positive charge from infinity to the given point against the electric forces."

Properties:

(i) Potential is a scalar quantity, its value may be positive, negative or zero.

(ii) S.I. Unit of potential is volt =  and its dimensional formula is [M1L2T-3I-1 ].

(iii) Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point.

(iv) Electric potential due to a positive charge is always positive and due to negative charge it is always negative except at infinity. (taking V¥ =0)

(v) Potential decreases in the direction of electric field.(a) Electric Potential due to a Point Charge in its Surrounding:

We know the region surrounding a charge is electric field. Thus we can also define electric potential in the surrounding of a point charge.

The potential at a point P at a distance x

from the charge q can be given as

Vp =

Where U is the potential energy of charge q0, if placed at point P, which can be given as

U =

Thus potential at point P is

The above result is valid only for electric potential in the surrounding of a point charge. If we wish to find electric potential in the surrounding of a charged extended body, we first find the potential due to an elemental charge dq on body by using the above result and then integrate the expression for the whole body.

(b) Electric Potential due to a Charge Rod:

Figure shows a charged rod of length L, uniformly charged with a charge Q. Due to this we will find electric potential at a point P at a distance r from one end of the rod shown in figure shown.

For this we consider an element of width dx at a distance x from the point P. Charge on this element is

The potential dV due to this element at point P can be given by using the result of a point charge as

Net electric potential at point P can be given as

(c) Electric Potential due to a Charged Ring:
Case I : At its centre

To find potential at the centre C of the ring, we first find potential dV at center due to an elemental charge dq on ring which is given as

As all dq's of the ring are situated at same distance R from the ring centre C, simply the potential due to all is added as being a scalar quantity, we can directly say that the electric potential at ring centre is . Here we can also state that even if charge Q is non-uniformly distributed on ring, the electric potential at C will remain same.

Case II : At a Point on Axis of Ring

If we wish to find the electric potential at a point P on the axis of ring as shown, we can directly state the result as here also all points of the ring are at the same distance

from the point P, thus the potential at P can be given as

Graph
(d) Electric Potential due to a Uniformly Charged Disc :

Figure shows a uniformly charged disc of radius R with surface charge density s C/m2. To find the electric potential at point P we consider an elemental ring of radius y and width dy, charge on this elemental ring is

dq = s. 2py dy

Due to this ring, the electric potential at point P can be given as

Net electric potential at point P due to whole disc can be given as

Net electric potential at point P due to whole disc can be given as

(e) Electric potential due to a closed disc at a point on the edge

Let us calculate the potential at the edge of a thin disc of radius 'R' carrying a uniformly distributed charge with surface density s.

Let AB be a diameter and A be a point where the potential is to be calculated. From A as centre, we draw two arcs of radii r and r dr as shown. The infinitesimal region between these two arcs is an element whose area is dA = (2rq) dr, where 2q is the angle subtended by this element PQ at point A. The potential at A due to the element PQ is

From D APB, we have

r = 2R cos q

or, dr = - 2R sin q dq

Hence

RELATION BETWEEN ELECTRIC FIELD INTENSITY AND ELECTRIC POTENTIAL

(a) For uniform electric field:

(i) Potential difference between two points A and B

(b) Non-uniform electric field:

where  = derivative of V with respect to x (keeping y and z constant)
= derivative of V with respect to y [keeping z and x constant]
= derivative of V with respect to z (keeping x and y constant)

(c) If electric potential and electric field depends only on one coordinate, say r:

(i)
is a unit vector along increasing r.
(ii)

is along the increasing direction of r.

(iii) The potential of a point

Area under E - x curve gives negative of change in potential. Negative of slope of V - x curve gives the electric field at that point.

ELECTRIC LINES OF FORCE
The idea of electric lines of force or the electric field lines introduced by Michael Faraday is a way to visualize electrostatic field geometrically.

The properties of electric lines of force are the following :

(i) The electric lines of force are continuous curves in an electric field starting from a positively charged body and ending on a negatively charged body.

(ii) The tangent to the curve at any point gives the direction of the electric field intensity at that point.

(iii) Electric lines of force never intersect since if they cross at a point, electric field intensity at the point will have two directions, which is not possible.

(iv) Electric lines of force do not pass but leave or end on a charged conductor normally. Suppose the lines of force are not perpendicular to the conductor surface. In this situation, the component of electric field parallel to the surface would cause the electrons to move and hence conductor will not remain equipotential which is absurd as in electrostatics conductor is an equipotential surface.

(v) The number of electric lines of force that originate from or terminate on a charge is proportional to the magnitude of the charge.

(vi) As a number of lines of force per unit area normal to the area at point represents the magnitude of intensity, crowded lines represent strong field while distant lines weak field. Further, if the lines of force are equidistant straight lines, the field is uniform

Electric lines offer due to two equal positive charges (field is zero at O). O is a null point
1. A charged particle need not follow a path.
2. Electric lines of force produced by static charges do not form close loop.

EQUIPOTENTIAL SURFACES

As shown in figure if a charge is shifted from a point A to B on a surface. M which is perpendicular to the direction of electric field, the work done in shifting will obviously, be zero as electric force is normal to the direction of displacement.

As no work is done in moving from A to B, we can say that A and B are at the same potential or we can say that all the points of surface M are at same potential or here we call surface M as equipotential surface.

Following figures show equipotential surfaces in the surrounding of point charge and a long charged wire

Every surface in the electric field in which at every point direction of electric field is normal to the surface can be regarded as equipotential surface.

Figure shows two equipotential surfaces in a uniform electric field E. If we wish to find the potential difference between two points A and B shown in figure, we simply find the potential difference between the two equipotential surfaces on which the points lie, given as

VA - VB = Ed

The figure shows a line charge with linear charge density l C/m. Here we wish to find potential difference between two points X and Y which lie on equipotential surfaces M1 & M2. To find the potential difference between these surfaces, we consider a point P at a distance x from wire as shown. The electric field at point P is

Now the potential difference between surface M1 and M2 can be given as

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