The Electric Field and Potential Difference

Electromagnetism Chapter I prof. Dr T. Fahmy

CHAPTER I

THE ELECTRIC FIELD AND POTENTIAL DIFFERENCE

1 Electromagnetism Chapter I prof. Dr T. Fahmy

After completing this chapter, the students should know:

 The different types of electric charges.

 The concept of Coulomb’s law.

 The relation between electric force and electric field.

 The units of electric charge, electric force and electric field.

 The concept of Gauss’s law.

 The relation between potential energy and potential difference.

 The motion of charged particle.

CHAPTER I THE ELECTRIC FIELD AND POTENTIAL DIFFERENCE

The Electrical Force Coulomb’s Law: Coulomb measured the magnitudes of the electric forces between charged objects using the torsion balance, and he showed that: 1- The electric force is inversely proportional to the square of the separation r between the particles and directed along the line joining them, i.e.  1 FC  2 (1.1) r

2- The electric force is directly proportional to the product of the charges q1 and

q2 on the particles.

 FC  q1 q2 (1.2)

Therefore, from these observations, we can express Coulomb’s law as follow:  q x q F  1 2 C r 2  q x q F  k 1 2 (1.3) C C r 2

Where kc is a constant called the Coulomb’s constant. The value of the

Coulomb’s constant depends on the choice of units. The Coulomb’s constant kc in SI units has the value of 1 kC  4o 9x109 N.m2 / C 2

- Where 0 is known as the permittivity of free space and has the value of 8.85x10 12 C2/N m2. The smallest unit of charge known in nature is the charge on the electron or proton, which has an absolute value of q e 1.6x1019 C Note that: 3 Electromagnetism Chapter I prof. Dr T. Fahmy

The electric force is attractive if the charges are of opposite sign and repulsive if the charges have the same sign.               (repulsion force)               (repulsion force)             (attraction force)

Example (1.1): The electron and proton of a hydrogen atom are separated by a distance of approximately 5.3x10-11 m. Find the magnitudes of the electric force and the

-31 gravitational force between the two particles, knowing that, me= 9.11x10 kg, mp=

-27 9 2 2 -11 2 2 1.67x10 , kc= 9x10 N.m /m and G= 6.7x10 N.m /kg . Solution: -e Firstly, the Coulomb force is  q x q P F  k 1 2 C C r 2 19 19 9 1.6x10 x1.6x10   9x10 x 2 5.3x1011  8.2 x108 N Secondly, the gravitational force is  m x m F G 1 2 G m r 2 31 27 11 9.11x10 x 1.67x10   6.7x10 x 2 5.3x1011  3.6 x1047 N

Example (1.2): Two charges, one of is +8x10−6 C and the other is −5x10−6 C, attract each other with a force of −40 N. How far apart are they? Solution: According to Coulomb’s law, we have  q x q 1 2 2 q1 x q2 FC  kC 2 r  k r C F

q x q 9x109 x8x106 x(5x106 ) r  k 1 2 r  C F (40)

r  9x103 m2  9x103 x106 mm2 30 10 mm

Example (1.3): Consider three point charges located at the corners of a right triangle as shown in figure, where q1=q3= 5 C and q2 = -2 C and a= 0.1 m.

F13y Find the resultant force exerted on q3. q2 F θ 23 Solution: F 0.1 m q 13x 3 Note that, the force F23 exerted q1 0.1 m by q2 on q3 is attractive F 13

Therefore, the magnitude of F23 is  q x q F  k 2 3 23 C a2 6 6 9 2x10 x 5x10  9x10 x     0.12 9 N

Also, the force F13 exerted by q1 on q3 is repulsive

Therefore, the magnitude of F13 is  q x q F  k 1 3 13 C ( 2 a)2 6 6 9 5x10 x 5x10   9x10 x 2  2 x 0.1 9 N

But F13 must be analyzed in two components in x-axis (horizontal component) and y- axis (vertical component), as follow:

 q x q F  k 1 3 cos,   450 13x C ( 2 a)2 6 6 9 5x10 x 5x10  1  9x10 x 2 x  2 x 0.1 2  7.9 N

 q x q F  k 1 3 sin,   450 13y C ( 2 a)2 6 6 9 5x10 x 5x10  1  9x10 x 2 x  2 x 0.1 2  7.9 N

Therefore, the total force in x-axis is

Fx= 7.9 – 9 = -1.1 N and the total force in y-axis is

Fy= 7.9 N

Then, the total force acting on the charge q3 is

   F total  F x  Fy

 F total (1.1)x  (7.9) y

The value of this forceis

2 2 F  (Fx )  (Fy )  (1.1)2  (7.9)2  7.97 N

Example (1.4): Consider three point charges located at the corners of a right triangle as shown in figure, where q1=2 C, q2= 5 C and q3 = 4 C. Find the resultant force exerted on q3.

F Solution: F 23y 13 q Note that, the force F13 exerted by q1 on q3 is repulsive 1 0.03 m q F 3 23x

6 0.04 m 0.05 m F 23

q 2 Electromagnetism Chapter I prof. Dr T. Fahmy

Therefore, the magnitude of F13 is

 q1 x q3 F13  kC 2 r13 2x106 x 4x106  9x109 x     0.032 80 N

Also, the force F23 exerted by q2 on q3 is repulsive

Therefore, the magnitude of F23 is

 q2 x q3 F23  kC 2 r23 5x106 x 4x106  9x109 x     0.052 72 N

But F23 must be analyzed in two components in x-axis (horizontal component) and y- axis (vertical component), as follow:  3 F  F cos, cos   23x 23 5 3  72 x  43.2 N 5

 4 F  F x sin, sin   23 y 23 5 4  72 x  57.6 N 5

Therefore, the total force in x-axis is

Fx= 43.2 + 80 = 123.2 N and the total force in y-axis is

Fy= 57.6 N

Then, the total force acting on the charge q3 is

   F total  F x  Fy

 F total (123.2)x  (57.6) y

The value of this forceis

2 2 F  (Fx )  (Fy )  (123.2)2  (57.6)2  136 N

The Electric Field The electric field E at a point in space is defined as ‘ the electric force F acting on a positive test charge q0 placed at that point divided by the magnitude of the test charge’’ E E Therefore, P P q   E  F (1.4) q0 q

To define the direction of an electric field, consider a point charge q located at a distance r from a test charge q0 located at a point p. According to Coulomb’s law, the electric force exerted by q on q0 is  q x q   k 0 (1.5) F C r 2 r Therefore   q   F  k (1.6) E C 2 r q0 r Also, note that: q At any point p the total electric field due to a group of charges3 equals the vector E 3 q E sum of the electric fields of the individual charges. 2 2 r Then: 3 r 2 q 1 r 8 4 r 1

r n q E n n Electromagnetism Chapter I prof. Dr T. Fahmy

                   E E1 E2 E3 E4 En    E n n1 q E1  q  4  k n (1.7) C  2 r n1 r n E 4 Example (1.5): Two point charges have the same charge q with opposite sign located at a distance r

  between them. Calculate the electric force F and electric field E on a third charge q3 if this charge is located at different positions such as a, b, c and d. Knowing that, q and q = 5 C, q = 4 C and r= 0.12 m. 1 2 3 d

c a b Solution: a At the point (a): ¼ r    3/4 r F a  F13  F23

q1 q3 q2 q3  kc x 2  kc x 2  3   1   r   r   4   4  6 6 6 6 9 5x10 x4x10 9 5x10 x4x10 9x10 2  9x10 2  3   1   x0.12  x0.12  4   4   22.22  200  222.22 N

  Fa 222.22 7 E a   6  5.55 x10 N / C q3 4x10

r r b At the point (b): There are two forces against to each other, therefore, we will obtain the total force as a difference between them as follow

   F b  F23  F13 q q q q k x 2 3 k x 1 3 c r2 c 2r2

5x106 x4x106 5x106 x4x106 9x109  9x109 0.122 2 x0.122  12.5  6.25  6.25 N   Fb 6.25 6 Eb   6  1.56 x10 N /C q3 4x10

½ r r At the point (c): There are two forces against to each other, therefore, we will obtain the total force as a difference between them as follow

   F c  F13  F23

q1 q3 q2 q3  kc x 2  kc x 2  1   3   r   r   2   2  6 6 6 6 9 5x10 x4x10 9 5x10 x4x10 9x10 2  9x10 2  1   3   x 0.12  x 0.12  2   2   50  5.55  44.45 N   Fc 6.25 7 E c   6  1.11x10 N / C q3 4x10

At the point (d) d

   F d  F13  F23

q1 q3 q2 q3  kc x 2  kc x 2 r r 2 Then  q q 5x106 x4x106 F  k x 1 3 9x109 x  12.5 N 13 c r2  0.122

 6 6 q2 q3 9 5x10 x4x10 F23  kc x 2  9x10 x 2  6.27 N r 2  0.12 2

Therefore   1 F  F x cos  6.27 x  4.43 N 23x 23 2   1 F  F x sin   6.27 x  4.43 N 23 y 23 2 Then,thetotal forceis    Fd  Fx  Fy  4.43x  12.5 4.43 y F  2  2  4.43 2  8.07 2  9.2 F x F y       Fc 9.2 6 E c   6  2.3 x10 N / C q3 4x10

Example (1.6): The electric field (E) in a certain neon sign is 5000 V/m. What force does this field (F) exert on a neon ion of mass 3.3x10−26 kg and charge +e? Solution:

The force (F) on the neon ion is

 F  q E

 F 1.6x1019 x5000 8x1016 N

Example (1.7): +Q =2C +Q =4C +Q =1C In the following figure calculate 1 2 3

r =2cm r =3cm 12 32

 1- The total force ( F ) acting on Q2, due to the effect Q1 and Q3

 2- The electric field ( E ) at the position of Q2. Solution: 1- The total force is

  =  + F F12 F32

Q xQ 2x106 x 4x106  K x 1 2 9x109 x F12 = c 2 = 2 2 r12  2x10 

8x1012 = 9x109 x 180 N (to the right) 4x104

Q x Q 5x106 x1x106  K x 3 2 9x109 x F13 = c 2 = 2 2 r32  3x10 

5x1012 = 9x109 x  50 N (to the left) 9x104 Therefore, the total force is

  =  + = 180 - 50 =130 N (to the right) F F12 F32

 2- The electric field E is

  F 130 = total = 3.25x107 N/m E  6 Q2 4x10

Example (1.8): -Q =2C +Q =4C +Q =5C In the following figure calculate 1 2 3

r =2cm r =3cm 12 32  3- The total force ( F ) acting on Q2, due to the effect Q1 and Q3

 4- The electric field ( E ) at the position of Q2. Solution: 3- The total force is

  =  + (to the left) F F12 F32 12 Electromagnetism Chapter I prof. Dr T. Fahmy

Q xQ 2x106 x 4x106  K x 1 2 9x109 x F12 = c 2 = 2 2 r12  2x10 

8x1012 = 9x109 x 180 N 4x104

Q x Q 5x106 x 4x106  K x 3 2 9x109 x F13 = c 2 = 2 2 r32  3x10 

20x1012 = 9x109 x  200 N 9x104 Therefore, the total force is

  =  + = 180 + 200 =380 N F F12 F32

 4- The electric field E is

  F 200 = total = 5x107 N/m E  6 Q2 4x10

Example (1.9):

Find the force on the charge q2 in the diagram below due to the charges q1 and q2.

q1 = 1 µC q2 = -2 µC q3 = 3 µC

0.1 m 0.15 m

Solution:

q3 = 3 µC Example (1.11):

0.15 m 13 q1 = 1 µC

q2 = -2 µC 0.1 m

Electromagnetism Chapter I prof. Dr T. Fahmy

Find the force on q2 in the diagram to the right.

Fx  F12  1.8 N Fy  F32  2.4 N 2 2 2 2 F2  Fx  Fy  (1.8 )  ( 2.4 )  3.0 N

Fy 2.4 tan    1.33 Fx 1.8    126.9o ( ccw from pos. x  axis )

q3 = 3 µC

F2 F32

q1 = 1 µC 

F12 q2 = -2 µC

Example (1.12):

Find the electric field at P due to charges q1 and q2. q2 = -1.5 µC

6 | q1 | 9 (2x10 ) 5 E1  ke 2  (8.99x10 ) 2 1.13x10 N /C r1 (0.4) | q | (1.5x106 ) 0.4 m P 2 9 4 q = 2 µC E2  ke 2  (8.99x10 ) 2  5.4x10 N /C 1 r2 (0.5) 5 4 Ex  E1  E2 cos 1.13x10  5.4x10 cos(36.9)  6.9x104 N /C 4 4 q2 = -1.5 µC E y  E2 sin  5.4x10 sin(36.9)  3.24x10 N /C 0.5 m 2 2 4 2 4 2 0.3 m E  Ex  E y  (6.9x10 )  (3.24x10 ) E2  7.6x104 N / C E  = 36.9o 1 1 o    tan (E y / Ex )  tan (3.24/ 6.9)  25 E 0.4 m P 1 q1 = 2 µC

Gauss's Law 14 Electromagnetism Chapter I prof. Dr T. Fahmy

This law is relating the distribution of electric charge to the resulting electric field. Gauss's law states that:

The electric flux through any closed surface is proportional to the enclosed electric charge.

The law was formulated by C. F. Gauss in 1835, but was not published until 1867. It is one of four of Maxwell's equations which form the basis of classical electrodynamics, the other three being Gauss's law for magnetism, Faraday's law of induction, and Ampère's law with Maxwell's correction. Gauss's law can be used to derive Coulomb's law, and vice versa. S E As shown in the figure, if a surface with an

Area of A is placed in an electric field of intensity

E, therefore, the total normal electric flux can be expressed (ɸ) as follow:

  E.S (1.8) r dS E So, the vertical flux (dɸ) through an area of (dS) can be written as follow: q

d  E.dS (1.9) where E is the electric field at any point of sphere surface and equals to

1 q E  2 4  o r

1 q d  2 dS 4  o r

Then by using the integration, the total flux can be determined:

2  1 q 4r   d  dS  2  0 4  o r 0

1 q 2   2 4 r 4 o r

q   (1.10) o

But if the surface of the sphere is non-homogeneous, the flux will be

  E S cos

   E S cos

Or in other form as follow:

   E S cos (1.12)

Therefore, from equations (1.11) and (1.12), we can conclude that:

1  E S cos   q (1.13) 0

Example (1.13):

Calculate the electric flux (φ) through a circle has a radius of 5 cm and a charge of 4 C at its center.

Solution: r = 5 cm φ= E. A

Q=4 r= 5x10-2 m and Q=4x10-6 C C

6 Q 9 4x10 6 E  kc 9x10 2 72x10 N / C r 2 5x102 

A  4 r 2  4x3.14 x(5x102 )2 3.14x102 m2

  72x106 x3.14x102  2.26x108 N.m2 / C

The Potential Energy (U) and Potential Difference (V)

If a test charge q0 is placed in an electric field E, created by some other charged objects, the electric force (F) acting on the test charge is q0 E. If the charge is moved a distance d ℓ , then the total work done can be written as follow:

dw F cos dℓ (1.14)

Where θ is the angle between the force (F) and the displacement (d ℓ). Therefore if the charge is moved from the point A to the point B, the total work done can be written as follow:

B wAB  F cos dℓ (1.15) A

Also, in terms of the potential energy (U), the equation (1.15) can be written as follow

B U   F cos dℓ AB  A

F  qo E

B U   q E cos dℓ (1.16) AB  0 A

The negative sign mean that, the motion against the electric field direction.

Note that,

0  at θ= 0 → WAB and UAB have a maximum value

0  at θ= 90 → WAB and UAB have a minimum value, i.e., (WAB and UAB = 0).

The Potential Difference (V)

The potential difference (V) is defined as the work done to transport a positive charge against the electric field.

The potential difference (V) is defined as the ratio between the potential energy

(U) to a test charge (q0).

Therefore,

U AB VAB  (1.17) q0

1 B V   q E cos dℓ AB  0 q0 A

B V   E cos dℓ AB  A

V   E cos dℓ AB 

If θ= 00

V   E dℓ AB 

Or in a differentiation form, we can write

dV   E dℓ dV E   (1.18) dℓ

Note that,

The potential difference is a scalar quantity.

Example (1.14):

Calculate the potential difference at a point located R at a distance r from the centre of a charged sphere with appositive charge? r

Solution:

As known,

q 1 E  2 (1) 40 r

dV E   dr dV   E dr (2)

From 1 and 2, we can write

q 1 dV   2 dr 4 o r

The value of V can be estimated using the integration process, as follow:

V q r 1 V  dV   dr   2 0 4 0 0 r q 1 V   K1 4 0 r

Where K1 is the integration constant. To calculate the value of this constant, the boundary condition must be used as follow;

at r → α →V=0 , therefore, K1= 0 ,

q 1 V  4 r o q V  k (3) r

Also, at the sphere surface, the potential difference is

q V  k (4) R

Example (1.15): c

In the following figure, 0.1 m 1- Calculate the potential at the points (a, b and c). 0.1 m

19 b a 0.04 m 0.06 m 0.04 m Electromagnetism Chapter I prof. Dr T. Fahmy

2- Calculate the potential energy of a positive

charge of 4x10-9 C at the points (a, b and c).

3- Calculate the potential difference Vab, Vba and Vcb.

4- Calculate the work to transfer a charge of 4x10-9 C

From a to b and from c to a.

Solution:

q1 q2 Va  k  k , q1  q2  q r1 r2

1-  1 1  9 6  1 1  Va  kq     9x10 x12x10    r1 r2  0.06 0.04 5 Va   9x10 V

9 6  1 1  4 Vb  9x10 x12x10   .25x10 V 0.04 0.14

9 6  1 1  Vc  9x10 x12x10   0 V 0.1 0.1

5 9 4 U a  qVa  9x10 x4x10  36x10 J 4 9 3 2- U b  qVb  225x10 x4x10  9x10 J 9 U c  qVc 0x4x10  0 J -

4 5 5 5 5 Vab  Vb  V a  225x10   9x0  22.5x10 9x10 31.5x10 V 5 4 5 5 5 3- Vba  Va  Vb   9x0  225x10   9x0  (22.5x10 )   31.5x10 V 4 Vcb  Vb  Vc  (225x10 ) V

Wab  Wb  Wa  qVab  4x109 x31.5x105  126x104 J 4- U ca Wa  Wc  qVca

 4x109 x 9x105   36x104 J

Example (1.16): The potential difference between a certain thundercloud and the ground is 7x106 V. Find the energy dissipated when a charge of 50 C is transferred from the cloud to the ground in a lightning stroke.

Solution: The energy is W= qV

= (50 C)(7 x106 V) 5x108

= 3. J

Note that:

We knew that, objects have potential energy because of their positions. In this case charge in an electric field has also potential energy because of its positions. Since there is a force on the charge and it does work against to this force we can say that it must have energy for doing work. In other words, we can say that Energy required increasing the distance between two charges to infinity or vice verse. Electric potential energy (U) is a scalar quantity and Joule is the unit of it. The following formula is used to calculate the magnitude of U;

Q x Q U  K x 1 2 c r

Be careful:  In this formula if the charges have opposite sign then, (U) becomes negative, if they are same type of charge then, (U) becomes positive.  If (U) is positive then, electric potential energy is inversely proportional to the distance (r).  If U is negative then, electric potential energy is directly proportional to the distance (r).

(a) (b) (c )

In Figure a and Figure b, charges repel each other, thus external forces does work for decreasing the distance between them. On the contrary, in Figure c, charges attract each other, distance between them is decreased by electric forces, and there is no need for other external forces. Q = 10q Q = 8q 1 4d 3

Example (1.17): System given below is composed of the charges 3d 10q, 8q and -5q. 5d Find the total electric potential energy of the system

Q = -5q Solution: 2 2 Q1 xQ2 10qx 5q  50 q  U12  Kc  Kc  Kc r12 3d 3d 2 Q1 xQ3 10qx8q 20 q  U13  Kc  Kc  Kc r13 4d d 2 Q2 xQ3  5qx8q  8q 

U23  Kc  Kc  Kc r23 5d d

Utotal U12 U13 U23  50 q2  20 q2   8q2   K  K  K c 3d c d c d 14 q2   K Joule c 3d

Note that:

Electric Potential Electric potential is the electric potential energy per unit charge. It is known as voltage in general, represented by V and has unit volt (Joule/C).

Electric Potential and Potential Energy due to Point Charge The potential difference V can be calculated between the points A and B as follow:

rB 22 Electromagnetism Chapter I prof. Dr T. Fahmy

rB V V   E dr B A  (1.19) rA But , as known q 1 r E  2 A 4 o r Therefore,

q rB 1 V V   dr B A 4  r 2 o rA q 1 rB   4 r o rA q  1 1      4 o rB rA 

If rA   1 q VB  4 o rB

or ingeneral form

1 q V  (1.20) 4o r Example (1.18):

Two charges of 2mC and -6mC are located at positions (0,0) m and (0,3) m, respectively as shown in figure 5.13. (i) Find the total electric potential due to these charges at point (4,0) m. (ii) How much work is required to bring a 3mC charge from  to the point P? (iii) What is the potential energy for the three charges?

q= -6 mC (0,3) Solution:

(i) (i) Vp = V1 + V2

1 q1 q2  V     q= 2 mC (4,0) 40  r1 r2  (0,0) p 23 Electromagnetism Chapter I prof. Dr T. Fahmy

6 6 9 2x10 6x10  3 V 9x10      6.3x10 volt  4 5 

(ii) The work required is given by

-6 3 -3 W = q3 Vp = 3  10  (-6.3  10 ) = -18.9  10 J

The -ve sign means that work is done by the charge for the movement from  to P.

(iii) The potential energy is given by

U = U12 + U13 + U23

q1 q2 q1 q3 q2 q3  U k      r12 r13 r23 

6 6 6 6 6 6 9 2x10  6x10  2x10 3x10   6x10 3x10  U 9x10      3 4 5 

U  5.5x102 J

Example (1.19):

A particle having a charge q=3x10-9C moves from point a to point b along a straight line, a total distance d=0.5m. The electric field is uniform along this line, in the direction from a to b, with magnitude E=200 N/C. Determine the force on q, the work done on it by the electric field, and the potential difference Va-Vb.

Solution:

The force is in the same direction as the electric field since the charge is positive; the magnitude of the force is given by

a F =qE = 3x10-9 x 200 = 600x10-9 N b d= 0.5 m

The work done by this force is

W =Fd = 600x10-9 x 0.5 = 300x10-9 J

The potential difference is the work per unit charge, which is

Va-Vb = W/q = 100 V

Va-Vb = Ed = 200 ´ 0.5 = 100 V

Example (1.20):

A charge q is distributed throughout a nonconducting spherical volume of radius R. (a) Show that the potential at a distance r from the center where r < R, is given by

q3R2 r 2  V  3 80 R

Solution:

To calculate the voltage inside the non-conducting spherical at a point A, we will calculate the voltage between a point at infinity and the point A as follow:

  V V  . A  E dℓ

Since, the field has two different values outside and inside the sphere as we know

q q r Eout  2 Ein  3 4or 4oR

VA V VA VB  VB V      V V   .   . A   Ein dℓ  Eout dℓ

Note that:

The angle between E and dℓ is 1800, i.e., cos 180= -1

Also dℓ=-dr

q r q V V   dr  dr A   3  2 4oR 4or q r 2  q 1 q 3R2  r 2       3     3 4oR  2  4o r  4oR

But if the point A at the surface of the sphere, the voltage will be

q V  4oR Example (1.21):

For the charge configuration shown in the following figure, Show that V(r) p for the points on the vertical axis, assuming r >> a, is given by

+q a +q a Solution: -q

Vp = V1 + V2 + V3

q q q V    40 (r  a) 40 r 40 (r  a)

q(r  a) q(r  a) q V  2 2  40 (r  a ) 40 r

2 a q q V    a2  4 r 4 r 2 1  0 0  2   r  when r>>a then a2/r2 <<1

2 a q q V    a2  4 r 4 r 2 1  0 0  2   r 

يمكن فك القوسين بنظرية ذات الحدين والحتفاظ بأول حدين فقط كتقريب جيد

(1 + x)n = 1 + nx when x<<1

2a q  a2  q V  1  2  2  40 r  r  40 r

ويمكن إهمال a2/r2 بالنسبة لـ 1

1  q 2 a q  V    2  40  r r 

Example (1.22):

Find the potential difference between points A and B, VAB in terms of kcq/r?

A Solution: 3r q W V V V  AB AB B A q K x q K x q V  c V  c A 3r B 2r

Kc x q  3 2  VAB VB VA     r  6 6  q 2r K xq B V  c AB 6r

The Motion of Charged Particle Under the Effect of Electric Field: If a negatively charged particle (the electron) moves under the effect of an electric field as shown in the figure, there are two forces acting on the electron as follow: +++++++++++++ F1  q E and e

F2  m a (1.21) Therefore q E  m a ------

Hence, the acceleration (1.22) (a)  q E m 27 Electromagnetism Chapter I prof. Dr T. Fahmy

The velocity of the charged particle can be estimated using Eq. (1.22) as follow q E a  m but, as known

dv a  dt Then dv q E q E   dv  dt dt m m v q E t q E v  dv  dt  v  t  K   1 0 m 0 m

Where K1 is the integration constant, and to calculate, the boundary conditions must be used as follow

At t=0 → y=0, then K2=0 then

q E v  t (1.23) m The displacement of the charged particle through the electric field can be deduced as follow q E v  t m but, as known

dy v  dt Then dy q E q E  t  dy  t dt dt m m y q E t q E y  dv  t dt  y  t 2  K   2 0 m 0 2m

Where K2 is the integration constant, and to calculate, the boundary conditions must be used as follow

At t=0 → v=0, then K2=0 then

q E y  t 2 (1.24) 2m In addition the kinetic energy of the charged particle can be expressed as follow: 1 K.E  mv2 2 But, using equation (1.24), we can get:

2 1 1  qE  K.E  mv2  m t  2 2  m 

1 q2E 2 K.E  t 2 (1.25) 2 m

Example (1.22): An electron is displaced a distance of 1 cm in an electric field with intensity of 104 N/C . Calculate the velocity of the electron, its kinetic energy and the required time to move this distance. Knowing that, e= 1.6x10-19 C and m= 9.11x10-31 Kg

Solution:

 The acceleration (a) is:

F e E 1.6x1019 x104 a     1.8x1015 m / sec2 m m 9.11x1031

 The velocity (v) can be calculated as follow:

v  2 a y  2 x1.8x1015 x 0.01  6x106 m / sec

 The kinetic energy (K.E) is:

1 1 2 K.E  mv2  x9.11x1031 x 6x106  16x1018 J 2 2

 The required time (t)

v 6x106 t   3.3x109 sec a 1.8x1015

Example (1.23): 29 Electromagnetism Chapter I prof. Dr T. Fahmy

The strength of the electric field in the region between two parallel deflecting plates of a cathode-ray oscilloscope is 25 kN/C. Determine the force exerted on an electron passing between these plates. What acceleration will the electron experience in this region?

Solution: d

As indicated in the figure, one of the plates is positively + - A charged, while the other is negatively charged. The electric + - a - b field between parallel charged plates is constant (as long as we + + - stay in the middle away from the edges of the plates), and is + - directed from the + plate toward the - plate. The electric force on a charge q placed in a field E is:

Fon q = q E

Since q is negative for an electron, then the force (F) on the electron is opposite to the direction of (E), i.e., to the left (toward the + plate). The magnitude of this force is:

-19 3 -15 |Fon e-| = e E = (1.6 x 10 )x(25 x 10 ) = 4 x 10 N While this is an extremely small force, we find the acceleration is quite large:

a = F/m = (4 x 10-15)/(9.11 x 10-31) = 4.4 x 1015 m/sec2.

The direction of the acceleration is, of course, the same as that of the force acting; toward the + plate.