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Electric Field Intensity

Electric Field Lines for a Positive and Negative Charge

It has been observed that problems of electro-static can be dealt with more easily if we introduce a field concept for electrostatic interactions. According to this concept, the two charges need to be in actual contact for the interactions between them to take place. The charges are capable of directly influencing the other charges placed at a distance, through the intervening medium.

The region, around any charge, in which its influence can be realized is known as the electric field of that charge.

The strength of an electric field is measured by the force experienced by a unit positive charge placed at that point. The direction of field is given by the direction of motion of a unit positive  charge if it were free to move.

The electric field \vec{E} produced by a charge Q at the location of a small “test” charge q is defined as the electric force \vec{F} exerted by Q on q, divided by the test charge q:


\vec{E} = \vec{F}/q

In magnitude E =F/q     ------(1)

SI Unit: Newton per coulomb (N/C)

Electric field of a point charge

Electric Field for a Point Charge

When a positive test charge is used, the electric field always has the same direction as the electric force on the test charge.

In accordance to Coulomb force the force between two charges Q and q is,

F = kQq/r2    ------(2)

Substitute equation (2) in equation (1), we get,

E =F/q 

= (kQq/r2)/q

= kQ/r2

So, E = kQ/r2

= (1/4πε0)(Q/r2)…… (3)

For several point charges q1, q2, …qn,at distances r1, r2,….,rn from Q, the resultant electric field will be,

E (r)=(1/4πε0) Σ(qi/ri 2)ri

Continuous charge distribution:

The charge is distributed continuously over some region, the sum becomes an integral.

So, E (r)=(1/4πε0)?(1/r2) \hat{r} dq

If the charge is spread out along a line, with charge-per-unit-length λ , then dq=(λ)dl

Similarly for surface charge σ,

dq=( σ)da


for volume charge ρ,

dq=( ρ )dv

Therefore electric field for line charge will be,

E (r)=(1/4πε0)? (λ (r')/r2) \hat{r} dl

For surface charge,

E (r)=(1/4πε0)? (σ (r')/r2) \hat{r} da

And for volume charge,

E (r)=(1/4πε0)? (ρ (r')/r2\hat{r} dv

Refer this simulation for electric field intensity

I use this animation as a lecture aide to introduce students to the schematic diagram and e-fields between a plate. Is shows two metal plates connected to a pair of batteries. you can click on/off the e-field and show a charged particle traveling across the plates. You can click a button labeled, "schematic" to show the symbols scientists and engineers use to illustrate the physical setup. Press, "Play," in the animation to see the charge accelerate across the plates.

Electric field of a line charge:-

Electric Field of a Line Charge

Fid the electric field a distance z above the midpoint of a straight line of length 2L, which carries a uniform line charge λ. From the figure, the horizontal components of the two fields cancel, and the net field of the pair is,

cosθ = z/r

r = (z2+x2)3/2


dq = λdx

dET = 2dE cosθ

= (2z/4πε0) (λ dx/r3)\hat{z}

So, dET = (2z/4πε0) (λ dx/(z2 +x2) 3/2) \hat{z}


E_{T} = \frac{2z\lambda }{4\pi \epsilon _{0}}\int_{0}^{L}\frac{dx}{\left ( z^{2}+x^{2} \right )^{3/2}}\hat{z}

Substitute x =z tanθ, so, dx= zsec2θ in the above equation, we get,

E_{T} = \frac{2z\lambda }{4\pi \epsilon _{0}z}\int_{0}^{L}cos\theta d\theta \hat{z}

=(2 λ /4πε0z) [L/(L2+z2)1/2]\hat{z}

So, ET = (1 /4πε0) [2λL/z(L2+z2)1/2]\hat{z}

Case -1

For points far from the line (z>>L), this result simplifies:

L2+z2 = z2

So, ET = (1 /4πε0) (2λL/z2) \hat{z}

= (1 /4πε0) (q/z2) \hat{z}    (Since, q=2λL)

Thus at large distance, the line looks like a point charge (, q=2λL). Therefore the field reduces to point charge q/4πε0 z2.

Case -2

In the limit L → ∞, we obtain the field of an infinite straight wire:

ET= (2λL)/(4πε0z)(L(1+(z2/L2))\hat{z}

=(2λL/4πε0zL )\hat{z}           (Since, L→ ∞, so z2/L2 =0)

=(2λ/4πε0z) \hat{z}

In general,

ET= (1/4πε0) (2λ/s)

Here, s is the distance from the wire.

Refer this video to more about on Electric Field


Electric lines of force

On account of electric intensity at all points in an electric field, unit positive charge will be urged to move in a definite direction when placed in the electric field. The path described by the unit positive charge is called the line of force.

An electric line of force is defined as the path, straight or curved, along which a unit positive charge is urged to move when free to do so in an electric field. The direction of motion of a unit positive charge gives the direction of line of force.

                                                  Electric Lines of Force

The lines of force are straight if the electric field is due to an isolated charge and are curved if the field is due to two or more charges placed near each other.

Thus, a line of force may also be defined as a curve, tangent at any point of which gives the direction of the electric intensity at that point.

  • Electric intensity, at any point, is due to the interaction of two electro-static fields; one due to source charge and other due to test charge.

  • A line of force starts from a positive charge and ends on a negative charge. In other words, line of force starts from higher potential and ends on lower potential.

  • Two lines of force never cross each other. If the two lines were to cross, two tangents could be drawn to the lines of force at the common point. This means that there could be two directions of intensity at a point which is impossible.

  • The lines of force meet the surface of a spherical conductor normally. If it were not so, the electric field will have a component parallel to the surface of the conductor. This would mean a flow of current which is absurd.

  • The lines of force never pass through the conductor. This explains the absence of electric field with in the conductor.

  • The lines of force tend to contract longitudinally or lengthwise. That is, they possess longitudinal tension. Due to this property the two unlike charges attract each other.

  • The lines of force tend to exert lateral (sideways) pressure. That is, they repel one another laterally. This explains the repulsion between two like charges.

Problem (JEE Advanced):

A spherical charged conductor has surface charge density σ. The electric field on its surface is E and electric potential of conductor is V. Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be,

(a) 2E, 2V               (b) 4E, 2V

(c) 4E, 4V               (d) 2E, 4V


We know that, the electric field for a spherical charged conductor will be, E = (1 /4πε0) (q/R2)

E∝1/R2 (q = constant)

Radius is halved. Therefore, electric field will become 4 times or 4E.

Further, V = (1 /4πε0) (q/R)∝1/R      (q = constant) 

Radius is halved, so potential will become two times or 2V.

From the above observation we conclude that, option (b) is correct.

Conceptual Questions


You could avoid the shock of static electricity by holding out a sharp needle as you reach for a metal doorknob or wall. How does that needle protect you from static electricity?


The needle emits charge into the air via a corona discharge.


The needle acts as your personal lightning rod. When you are carrying a net electric charge, some of that charge settles onto the needle. The strong electric field near the needle’s tip initiates a corona discharge, and much of your accumulated charge leaves through it. This discharge limits your net electric charge and thus the size of any shock you experience.



A thick wire connects the lightning rod on the courthouse steeple to the ground and normally ensures that the rod is electrically neutral. However, when a negatively charged cloud floats overhead, what charge does that rod acquire?


The rod becomes positively charged.


The lightning rod charges by induction—the cloud’s negative charge attracts positive charge up the wire and onto the rod. The rod’s sharp point initiates a corona discharge, spraying positive charge toward the cloud and gradually decreasing the cloud’s charge. In that fashion, the lightning rod acts to suppress lightning strikes.



If you remove just one of the wires from an automobile’s battery, the vehicle will not start at all. Why doesn’t the other wire supply any energy?


Removing a single wire from the battery breaks the circuit and prevents a steady flow of electric current through the car’s electric system.


Neither the battery nor the rest of the car can accumulate charges indefinitely. Without one wire to carry charges from the battery to the car and a second wire to return those charges from the car to the battery, accumulation will occur and charge movement will stop.



A walk across wool carpeting in rubber-soled shoes has left you covered with negative charges. If you bring your hand near a large piece of metal, the negative charges will leap through the air as a spark in order to reach the metal. Which way is current flowingin this spark?


The current flows from the metal toward your hand.


Because current is defined as the flow of positive charges, it points in the direction opposite the flow of negative charges. Thus the current flows from the metal toward your hand. These charges move only briefly because there is no circuit. For the charges to move continuously they would have to be recycled and a circuit would be essential.

Question 1

A negatively charged rod is held near, but does not touch the knob of an electroscope. The leaves of the electroscope move apart from one another. A wire is connected to the knob and to a water tap with the negatively charged rod staying in the same position. Which of the following would occur?

(a) Electrons flow from the earth through the wire to the electroscope.

(b) Electrons flow from the electroscope through the wire to the earth.

(c) The leaves of the electroscope move closer together.

(d) The leaves of the electroscope remain still.

Question 2

(a) The law of electric charges states that opposite charges

(b) attract each other, similar charges attract neutral objects, and charged objects repel one another

(c) repel each other, similar charges attract neutral objects, and charged objects attract one another

(d) attract neutral objects, similar charges repel each other, and charged objects attract one another

(e) attract each other, similar charges repel one another, and charged objects attract some neutral objects

Question 3

The electric field 0.50 m from a small sphere with a positive charge of 7.2 \times10–5 C is

(a) 2.6 \times106 N/C [inward]              (b) 2.9 \times10–4 N/C [outward]

(c) 2.6 \times106 N/C [outward]          (d) 2.9 \times10–4 N/C [inward]

Question 4

The magnitude of the electric field between the plates of a parallel plate capacitor is 4.7\times104 N/C. If the charge on each plate were to increase by a factor of three, the magnitude of the electric field would

(a) increase by a factor of nine          (b) increase by a factor of three

(c) decrease by a factor of nine        (d) decrease by a factor of nine

Question 5

The magnitude of the electric field between the plates of a parallel plate capacitor is 4.7 \times 104 N/C. If the plates were separated to a distance that is twice their original separation distance, the magnitude of the electric field would

(a) double                                          (b) be halved

(c) decrease by a factor of sixteen       (d) not affected

Q.1 Q.2 Q.3 Q.4 Q.5






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