Chapter-21
Electric Charge and Electric Field
Topic Objectives:
The topic provides the basic concepts and terminology for understanding of electric charge, electrostatic
induction, Coulomb’s law and electric dipole. Of particular importance are the concepts of electrostatic
induction, Coulomb’s law and electric dipole etc. When studying this topic, please pay attention to (i)
Terminology (ii) Typical value of laws (iii) All boxed equations.
Contents:
Electric Charge
Conductor , Insulator and Induced charges
Coulomb’s Law
Electric Field and electric Forces
Electric Dipoles
Test Your Understanding (TYU)
1. Test Your Understanding of Section 21.3 Two equal positive charges = = 2 μ are
located at = 0, = 0.3 and = 0, = −0.3 , respectively. The magnitude and
direction of the total electric force that and exert on a third charge = 4μ =
0.4 , = 0 can be calculated. Suppose that charge = were -2µC. In this case, the
total electric force on Q would be (i) in the positive x-direction; (ii) in the negative x-
direction; (iii) in the positive y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of
these.
Ans: (iv) The x-components of the two forces cancel to each other while the (negative) y-
components add together, and the total electric force is in the negative y-direction
2. Test Your Understanding of Section 21.4 (a) A negative point charge moves along a
straight-line path directly toward a stationary positive point charge. Which aspect(s) of the
electric force on the negative point charge will remain constant as it moves? (i) magnitude;
(ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor direction. (b) A
negative point charge moves along a circular orbit around a positive point charge. Which
aspect(s) of the electric force on the negative point charge will remain constant as it moves?
(i) magnitude; (ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor
direction.
Ans: (a) (ii) As electric force is inversely proportional to the square of the distance r from
the charge to the field point. So magnitude of electric force increases as
distance decreases but direction remains constant. (b) (i) A negative point charge moves
along a circular orbit around a positive point charge, as distance r is constant so magnitude
of electric force remains constant and direction changes continuously.
, In Class Problem (ICP)
1. In class problem: (refer text book-21.4): Two equal positive charges = = 2 μ are
located at = 0, = 0.3 and = 0, = −0.3 , respectively. What are the magnitude
and direction of the total electric force that and exert on a third charge =
4μ = 0.4 , = 0?
Indentify and set up:
= 2μ = 2 × 10 , ( , ) = (0, ±0.3), = 4 μ = 4 × 10 , ( , ) = (0.4, 0)
× × × × × .
Execute: = cosα = ( . )
× = 0.23N
.
1 q Q 9 × 10 × 2 × 10 × 4 × 10 0.4
= cosα = × = 0.23N
4πϵ r (0.5) 0.5
From symmetry we see that the y-components of the two forces are equal and opposite. Hence
their sum is zero and the total force on Q has only an x-component = +
= 0.23 + 0.23 = 0.46 . The total force on Q is in the +x-direction, with
magnitude 0.46 N.
2. In class problem: (refer text book-21.9): Charge Q is uniformly distributed around a
conducting ring of radius a. Find the electric field at a point P on the ring axis at a distance x
from its center.
Indentify and set up: This is a problem in the superposition of electric fields. Each bit of
charge around the ring produces an electric field at an arbitrary point on the x-axis.
Execute: = ( )
=
1
=
4 ( + )( + )
, 1
= ⁄
∵ = = ℎ
4 ( + )
= = ( ) ⁄
= ( ) ⁄
×2
( × )
= ( ) ⁄
= ( ) ⁄
= ̂= ( ) ⁄
̂
3. In class problem: (refer text book-21.13): Figure shows an
electric dipole in a uniform electric field of magnitude 5 ×
105 N/C that is directed parallel to the plane of the figure. The
charges are ±1.6 × 10 ; both lie in the plane and are
separated by 0.125 nm = 0.125 × 10-9 m. find (a) the net force
exerted by the field on the dipole; (b) the magnitude and
direction of the dipole moment; (c) the magnitude and
direction of the torque; (d) the potential energy of the system in the position shown.
Indentify and set up:
This problem uses the ideas of this section about an electric dipole placed in an electric field.
= 1.6 × 10 , = 0.125 × 10 = 5 × 10 /
Execute:
(a) The field is uniform, so the forces on the two charges are equal and opposite. Hence the
total force on the dipole is zero.
(b) = = 1.6 × 10 × 0.125 × 10 = 2 × 10 C. m
The direction of is from the negative to the positive charge, 1450 clockwise from the
electric- field direction
(c) = = 2 × 10 × 5 × 10 × sin(145 ) = 5.7 × 10 N. m
and the direction of the torque is into the plane of paper.
(d) =− = − 2 × 10 × 5 × 10 × cos(145 ) = 8.2 × 10 J
, Assignment problem
1. Assignment problem (Ref. text book-21.7) Consider an electron-proton pair and compare its
electrostatic force with that of gravitational force ( = 6.67 × 10 Nm2/kg2)
Indentify and set up:
= 9.1 × 10 , = 1.67 × 10 ,
= 6.67 × 10 Nm /kg = ±1.6 × 10−9
Execute:
× × . × × . × . ×
= = =
6.67 × 10−11 × . × × . × . ×
= = =
. ×
= . ×
= 2.27 × 10
2. Assignment problem (Ref. text book-21.43) Two positive point charges q are placed on the
x-axis, one at x = a and one at x = -a. (a) Find the magnitude and direction of the electric field
at x = 0. (b) Derive an expression for the electric field at points on the x-axis. Use your result
to graph the x-component of the electric field as a function of x, for values of x between -4a
and +4a.
Indentify and set up:
Execute: (a) the resultant electric field E at x = 0 will be zero because of charge and
distance are same from origin. = + = ̂+ (− ̂) = 0
(b) (i) When the test charge q is placed at R in between –a to a.
=− ( )
and = ( )
= − =− ( )
+( )
=− ( )
(ii) When the test charge q is placed at S in between –a to a.
= ( )
and = ( )
= + = ( )
+( )
= ( )
Electric Charge and Electric Field
Topic Objectives:
The topic provides the basic concepts and terminology for understanding of electric charge, electrostatic
induction, Coulomb’s law and electric dipole. Of particular importance are the concepts of electrostatic
induction, Coulomb’s law and electric dipole etc. When studying this topic, please pay attention to (i)
Terminology (ii) Typical value of laws (iii) All boxed equations.
Contents:
Electric Charge
Conductor , Insulator and Induced charges
Coulomb’s Law
Electric Field and electric Forces
Electric Dipoles
Test Your Understanding (TYU)
1. Test Your Understanding of Section 21.3 Two equal positive charges = = 2 μ are
located at = 0, = 0.3 and = 0, = −0.3 , respectively. The magnitude and
direction of the total electric force that and exert on a third charge = 4μ =
0.4 , = 0 can be calculated. Suppose that charge = were -2µC. In this case, the
total electric force on Q would be (i) in the positive x-direction; (ii) in the negative x-
direction; (iii) in the positive y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of
these.
Ans: (iv) The x-components of the two forces cancel to each other while the (negative) y-
components add together, and the total electric force is in the negative y-direction
2. Test Your Understanding of Section 21.4 (a) A negative point charge moves along a
straight-line path directly toward a stationary positive point charge. Which aspect(s) of the
electric force on the negative point charge will remain constant as it moves? (i) magnitude;
(ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor direction. (b) A
negative point charge moves along a circular orbit around a positive point charge. Which
aspect(s) of the electric force on the negative point charge will remain constant as it moves?
(i) magnitude; (ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor
direction.
Ans: (a) (ii) As electric force is inversely proportional to the square of the distance r from
the charge to the field point. So magnitude of electric force increases as
distance decreases but direction remains constant. (b) (i) A negative point charge moves
along a circular orbit around a positive point charge, as distance r is constant so magnitude
of electric force remains constant and direction changes continuously.
, In Class Problem (ICP)
1. In class problem: (refer text book-21.4): Two equal positive charges = = 2 μ are
located at = 0, = 0.3 and = 0, = −0.3 , respectively. What are the magnitude
and direction of the total electric force that and exert on a third charge =
4μ = 0.4 , = 0?
Indentify and set up:
= 2μ = 2 × 10 , ( , ) = (0, ±0.3), = 4 μ = 4 × 10 , ( , ) = (0.4, 0)
× × × × × .
Execute: = cosα = ( . )
× = 0.23N
.
1 q Q 9 × 10 × 2 × 10 × 4 × 10 0.4
= cosα = × = 0.23N
4πϵ r (0.5) 0.5
From symmetry we see that the y-components of the two forces are equal and opposite. Hence
their sum is zero and the total force on Q has only an x-component = +
= 0.23 + 0.23 = 0.46 . The total force on Q is in the +x-direction, with
magnitude 0.46 N.
2. In class problem: (refer text book-21.9): Charge Q is uniformly distributed around a
conducting ring of radius a. Find the electric field at a point P on the ring axis at a distance x
from its center.
Indentify and set up: This is a problem in the superposition of electric fields. Each bit of
charge around the ring produces an electric field at an arbitrary point on the x-axis.
Execute: = ( )
=
1
=
4 ( + )( + )
, 1
= ⁄
∵ = = ℎ
4 ( + )
= = ( ) ⁄
= ( ) ⁄
×2
( × )
= ( ) ⁄
= ( ) ⁄
= ̂= ( ) ⁄
̂
3. In class problem: (refer text book-21.13): Figure shows an
electric dipole in a uniform electric field of magnitude 5 ×
105 N/C that is directed parallel to the plane of the figure. The
charges are ±1.6 × 10 ; both lie in the plane and are
separated by 0.125 nm = 0.125 × 10-9 m. find (a) the net force
exerted by the field on the dipole; (b) the magnitude and
direction of the dipole moment; (c) the magnitude and
direction of the torque; (d) the potential energy of the system in the position shown.
Indentify and set up:
This problem uses the ideas of this section about an electric dipole placed in an electric field.
= 1.6 × 10 , = 0.125 × 10 = 5 × 10 /
Execute:
(a) The field is uniform, so the forces on the two charges are equal and opposite. Hence the
total force on the dipole is zero.
(b) = = 1.6 × 10 × 0.125 × 10 = 2 × 10 C. m
The direction of is from the negative to the positive charge, 1450 clockwise from the
electric- field direction
(c) = = 2 × 10 × 5 × 10 × sin(145 ) = 5.7 × 10 N. m
and the direction of the torque is into the plane of paper.
(d) =− = − 2 × 10 × 5 × 10 × cos(145 ) = 8.2 × 10 J
, Assignment problem
1. Assignment problem (Ref. text book-21.7) Consider an electron-proton pair and compare its
electrostatic force with that of gravitational force ( = 6.67 × 10 Nm2/kg2)
Indentify and set up:
= 9.1 × 10 , = 1.67 × 10 ,
= 6.67 × 10 Nm /kg = ±1.6 × 10−9
Execute:
× × . × × . × . ×
= = =
6.67 × 10−11 × . × × . × . ×
= = =
. ×
= . ×
= 2.27 × 10
2. Assignment problem (Ref. text book-21.43) Two positive point charges q are placed on the
x-axis, one at x = a and one at x = -a. (a) Find the magnitude and direction of the electric field
at x = 0. (b) Derive an expression for the electric field at points on the x-axis. Use your result
to graph the x-component of the electric field as a function of x, for values of x between -4a
and +4a.
Indentify and set up:
Execute: (a) the resultant electric field E at x = 0 will be zero because of charge and
distance are same from origin. = + = ̂+ (− ̂) = 0
(b) (i) When the test charge q is placed at R in between –a to a.
=− ( )
and = ( )
= − =− ( )
+( )
=− ( )
(ii) When the test charge q is placed at S in between –a to a.
= ( )
and = ( )
= + = ( )
+( )
= ( )
