Electric Field Intensity

 Electric Field Intensity

AU : May-06,07,10,12,15,16,19, Dec.-03,09,12,17,18

• Consider a point charge Q₁ as shown in Fig. 2.3.1 (a).

• If any other similar charge Q₂ is brought near it, Q₂ experiences a force. Infact if Q₂ is moved around Q₁, still Q₂ experiences a force as shown in the Fig. 2.3.1 (a).

• Thus there exists a region around a charge in which it exerts a force on any other charge. This region where a particular charge exerts a force on any other charge located in that region is called electric field of that charge. The electric field of Q₁ is shown in the Fig. 2.3.1 (b).

• The force experienced by the charge Q₂ due to Q₁ is given by Coulomb's law as,

• Thus force per unit charge can be written as,

• This force exerted per unit charge is called electric field intensity or electric field strength. It is a vector quantity and is directed along a segment from the charge Q₁ to the position of any other charge. It is denoted as .

where p = Position of any other charge around Q₁

• The equation (2.3.2) is the electric field intensity due to a single point charge Q₁ in a free space or vacuum.

• Another definition of electric field intensity is the force experienced by a unit positive test charge i.e. Q₂ = 1C

• Consider a charge Q₁ as shown in the Fig. 2.3.2. The unit positive charge Q₂ = 1 C is placed at a distance R from Q₁. Then the force acting on Q₂ due to Q₁ is along the unit vector . As the charge Q₂ is unit charge, the force exerted on Q₂ is nothing but electric field intensity  of Q₁ at the point where unit charge is placed.

 

• If a charge Q₁ is located at the center of the spherical co-ordinate system then unit vector  in the equation (2.3.3) becomes the radial unit vector  coming radially outwards from Q₁. And the distance R is the radius of the sphere r.

 

 

1. Units of 

• The definition of electric field intensity is,

 = Force/Unite Charge = (N) Newtons/(C) Coulomb

• Hence units of  is N/C. But the electric potential has units J/C i.e. Nm/C and hence  is also measured in units V/m (volts per metre). This unit is used practically to express .

 

2. Method of Obtaining E in Cartesian System

• Consider a charge Qi located at point A(x₁,y₁,z₁) as shown in the Fig. 2.3.3. It is required to obtain  at any point B (x, y, z) in the cartesian system. Then  at point B can be obtained using following steps :

 

Step 1 : Obtain the position vectors of points A and B.

• Substituting  interms of the cartesian co-ordinates of A and B, the required electric field intensity  at the point B can be obtained.

 

3. Electric Field due to Discrete Charges

• Similar to a force exerted on a charge due to n number of charges is the vector sum of the individual force exerted by each charge, the electric field at a point due to n number of charges is to be obtained using law of superposition.

• Consider n charges Q₁,Q₂ ...Q_n as shown in the Fig. 2.3.4. The combined electric field intensity is to be obtained at point P. The distances of point P from Q₁,Q₂ ...Q_n are R₁,R₂,...R_n respectively. The unit vectors along these directions are a  respectively.

• Then the total electric field intensity at point P is the vector sum of the individual field intensities produced by the various charges at the point P.

• Each unit vector can be obtained by using the method discussed earlier.

 

4. Important Observations

• The important observations related to  are,

1.  around a charge Q₁ is directly proportional to the charge Q₁.

2.  around a charge Q₁ is inversely proportional to the distance between charge Q₁ and point at which  is to be calculated. More is the distance less is the electric field intensity and less will be the force experience by a unit charge placed at that point.

3. The field intensity  at any point and force  exerted on a charge placed at the same point are always in the same direction.

4. Placing unit charge is a method of detecting the presence of electric field around a charge. Without any unit test charge placed nearby, every charge has its electric field always existing, around it.

5. The test charge placed must be small enough so that the electric field intensity  to be measured should not get disturbed.

 

Ex. 2.3.1 A charge of - 0.3 ,µC is located at A(25, - 30, 25) (in cm) and a second charge of 0.5µC is at B (-10, 8,12) cm. Find the electric field intensity at the origin. AU : May-10, Dec.-03, 17, Marks 4

Sol. : A (0.25, - 0.3, 0.15) m, - 0.3 µC

B (- 0.1, 0.08, 0.12) m, 0.5 µC

O (0, 0, 0)

 

Ex. 2.3.2 Consider a square of side 5 cm. Three positive charges of 100 nC each are located at three comers of the square. Find the value of the electric field intensity at the fourth corner of the square.AU : May-12, Marks 8

Sol. : The arrangement is shown in the Fig. 2.3.6. Let the charges are placed at A, B and C while   is to be obtained at fourth comer O.

 

Ex. 2.3.3 Point charges 5 nC and - 2 nC are located at (2, 0, 4) and (- 3, 0, 5), respectively. 1) Determine the force on a 1 nC point charge located at (1, - 3, 7), 2) Find the electric field intensity at (1, - 3, 7)AU: Dec.-12,18, Marks 8

Sol : The arrangement is shown in the Fig. 2.3.7.

Ex. 2.3.4 Three point charges in free space are located as fllows 50 nC at (0,0) m; 40 nC at (3,0) m; = 60 nC at (0,4) m. Find the electric field intensity at (3,4) m.

Sol : The charges are shown in the Fig. 2.3.8.

Ex. 2.3.5 A positive point charge 100 × 10^-12 C is located in air at x = 0, y = 0.1 m and another such charge at x = 0, y = -.01 m. What is the magnitude and direction of E at x = 0.2 m , y = 0.m AU:  May-15, Marks 6

So. : The charges are shown in the Fig. 2.3.9

 

Ex. 2.3.6 A charge of 1 C is at (2, 0, 0). What charge must be placed at (- 2, 0, 0) which will make y component of total   zero at the point (1, 2, 2) ? AU : Hay-19, Marks 13

Sol. : The various points and charges are shown in the Fig. 2.3.10.

The position vectors of points A, B and P are,

This is the required charge Q₂ to be placed at (-2,0,0) which will make y component of   zero at point P.

Examples for Practice

Ex. 2.3.7 Determine electric field   at the origin due to a point charge of 54.9 nC located at (- 4, 5, 3) m in cartesian co-ordinates.

 

Ex. 2.3.8 Calculate  at P(l, 1, 1) in free space caused by four identical 3 nC point charges located at P₁ = (1, 1, 0), P₂ = (-1, 1, 0), P₃ = (-1, -1, 0) and P₄ = (1, -1, 0).

 

Ex. 2.3.9 A charge Q₁ = - 10 nC is at the origin in free space. If the x-component of E is to be zero, at the point (3, 1, 1), what charge Q_t should be kept at the point (2, 0, 0) ? [Ans.: 4.2728 nC]

Review Questions

1. Define electric field intensity and explain the procedure to obtain electric field at a point due to number of discrete charges. 

AU : May-07, Dec.-09, Marks 8

2. Derive electric field intensity at a point in a field due to a point charge.

AU : May-06, Marks 6