Semester 2 Physics (SF 026) Lecture: BP 3 by Yew Sze Fiona Website:

Semester 2 Physics (SF 026) Lecture: BP 3 by Yew Sze Ling @ Fiona Website: http://yslphysics.weebly.com/

Chapter 1: Electrostatics The study of electric charges at rest, the forces between them and the electric fields associated with them.

Overview Electrostatic Coulomb s Law F Qq 2 r Electric Field E F q Electric Potential V W q o Electric Field For Point Charge E kq 2 r Charge in a Uniform Electric Field E V d Equipotential Surface Electric Potential Energy U kqq 2 r

1.1 Coulomb s Law State Coulomb s law, F Qq r kqq 2 2 4 o r Sketch the electric force diagram and apply Coulomb s law for a system of point charges. (2D, maximum four charges) Learning Objectives

Coulomb s Law Coulomb s law states that the magnitude of the electrostatic (Coulomb/electric) force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Qq F 2 r Coulomb s Law Equation Electrostatic Force

Coulomb s Law Mathematically, F kqq r 2 Qq 4 o r 2 where F : magnitude of electrostatic (Coulomb s) force Q, q : magnitude of charges r : distance between two point charges k : electrostatic constant, ε o k = 9.0 10 9 N m 2 C -2 : permittivity of free space, ε o = 8.85 10-12 C 2 N -1 m -2 k 1 4 0

Coulomb s Law Graphically, F F Gradient, m = kqq 1 0 r 0 r 2 Notes: o The sign of the charge can be ignored when substituting into the Coulomb s law equation. o The sign of the charges is important in distinguishing the direction of the electric force.

Electric Force Diagram There are two types of charges in nature positive and negative charges. Like charges repel Repulsive force Unlike charges attract Attractive force The direction of the force is along the straight line joining the two point charges.

Example 1 Sketch the force diagram for q 1. a) b) F 21 + F 31 q 1 q 2 q 3 + q 2 + q 1 F 31 F 21 q 3

Example 2 Two point charges, q 1 = 20 nc and q 2 = 90 nc, are separated by a distance of 4.0 cm as shown in figure below. Find the magnitude and direction of a. the electric force that q 1 exerts on q 2. b. the electric force that q 2 exerts on q 1. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 2 Solution

Example 3 Three point charges lie along the x-axis as shown in figure below. Calculate the magnitude and direction of the total electric force exerted on q 2. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 3 Solution

Example 4 Figure below shows the three point charges are placed in the shape of triangular. Determine the magnitude and direction of the resultant electric force exerted on q 1. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 4 Solution

Example 5 Two point charges, q 1 = +4.0µC and q 2 = +6.0µC, are separated by a distance of 50 cm as shown in figure below. Determine the position of a point charge q that is placed on the line joining q 1 and q 2 such that the net force acting on it is zero. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 5 Solution

1.2 Electric Field Define and use electric field strength, Use E kq 2 r E F q for point charge. Sketch the electric field strength diagram and determine electric field strength E for a system of charges. (2D, maximum four charges) Learning Objectives

Electric Field Electric field is defined as a region of space around isolated charge where an electric force is experienced if a (positive) test charge is placed in the region. + E + Electric field around charges can be represented by drawing a series of lines. These lines are called electric field lines (lines of force). The direction of electric field is tangent to the electric field line at each point.

Electric Field Lines (a) Isolated point charge Single positive charge Single negative charge The lines point radially outward from the charge The lines point radially inward from the charge

Electric Field Lines (b) Two charges Two equal point charges of opposite sign, +q and -q The lines are curved and they are directed from the positive charge to the negative charge.

Electric Field Lines Two equal positive charges, +q and +q Neutral point is defined as a point (region) where the total electric force is zero. It lies along the vertical dash line.

Electric Field Lines Two opposite unequal charges, +2q and -q note that twice as many lines leave +2q as there are lines entering q number of lines is proportional to magnitude of charge.

Electric Field Lines (c) Two opposite charged parallel metal plate The lines go directly from positive plate to the negative plate. The field lines are parallel and equally spaced in the central region far from the edges but fringe outward near the edges. Thus, in the central region, the electric field has the same magnitude at all points. The fringing of the field near the edges can be ignored because the separation of the plates is small compared to their size.

Electric Field Lines Characteristic of electric field lines: The field lines indicate the direction of the electric field (the field points in the direction tangent to the field line at any point). The lines are drawn so that the magnitude of electric field is proportional to the number of lines crossing unit area perpendicular to the lines. The closer the lines, the stronger the field. Electric field lines start on positive charges and end on negative charges, and the number starting or ending is proportional to the magnitude of the charge. The field lines never cross because the electric field don t have two value at the same point.

Example 6 Sketch electric field lines for the diagram below: a) b)

Electric Field Strength The electric field strength at a point is defined as the electric (electrostatic) force per unit (positive) test charge. Mathematically where, E E : magnitude of the electric field strength F : magnitude of the electric force q 0 : magnitude of test charge F q 0

Electric Field Strength Since, F kqq 2 r thus, E kqq 2 r q It is a vector quantity. E kq 2 r In the calculation of magnitude E, substitute the MAGNITUDE of the charge only. The units of electric field strength is N C -1 or V m -1. The direction of the electric field strength, E depends on the sign of isolated charge.

What is the difference between isolated charge and test charge? E F q 0 E k Q 2 r Isolated charge Test charge + Test charge + E Isolated charge

Since both F and E are vectors, how to determine the direction of F and E for a test charge?? The direction of electric field strength, E depends on sign of isolated point charge. The direction of the electric force, F depends on the sign of isolated point charge and test charge.

A positive isolated point charge A negative isolated point charge

Example 7 A metal sphere can be regarded as a point object in space. It carries a charge of +6.0 µc. Find the electric field that the sphere generates at a distance of 10 cm around it. Solution

Example 8 Two point charges, q 1 = 1 C and q 2 = 4 C, are placed 2 cm and 3 cm from the point A respectively as shown in figure below. Find a) the magnitude and direction of the electric field intensity at point A. b) the total electric force exerted on q 0 = 4 C if it is placed at point A. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 8 Solution

Example 9 Two point charges, q 1 = 12 nc and q 2 = 12 nc, are placed 0.10 m apart. Calculate the total electric field at point a and point b.

Example 9 Solution

1.3 Electric Potential Define electric potential, V W q o Define and sketch equipotential lines and surfaces of an isolated charge and a uniform electric field Use V kq r system of charges. for a point charge and a Learning Objectives

1.3 Electric Potential Calculate potential difference between two points V V final V initial Deduce the change in potential energy between two points in electric field U qv Calculate potential energy of a system of point charges U q1q k r12 2 q1q r 13 3 q2q r 23 3 W q o Learning Objectives

Electric Potential Electric potential, V of a point in the electric field is defined as the work done in bringing (positive) test charge from infinity to that point in the electric field per unit test charge. V W q o W : Work done q o : Test charge

Electric Potential OR Electric potential, V of a point in an electric field is defined as the potential energy per unit positive charge at that point in the electric field. V U q o U : Potential energy q o : Test charge It is a scalar quantity. The unit of electric potential is Volt (V) OR J C 1. The electric potential at infinity is considered zero. (V = 0)

Extra Note Q + A F ext F E r +q o r = dw dw xr x dw where F F F ext E dr E xr dr x F E dr kqq 2 r o W W W kqq kqq kqq r o o r o r r 2 1 dr r V V V A A A W q o kqq rq kq r r The electric potential at point A at distance r from a positive point charge Q o o

Electric Potential Since W kqq r o thus the equation of electric potential can also be written as V kq r

Electric Potential U=q o V The electric potential energy of a positively charged particle increases when it moves to a point of higher potential. The electric potential energy of a negatively charged particle increases when it moves to a point of lower potential. If the value of work done is negative work done by the electric force (system). If the value of work done is positive work done by the external force or on the system. In the calculation of U, W and V, the sign of the charge MUST be substituted in the related equations.

Equipotential Surface Equipotential surface (line)is defined as the locus of points that have the same electric potential. A point charge A uniform electric field

Equipotential Surface The dashed lines represent the equipotential surface (line). The equipotential surfaces (lines) always perpendicular to the electric field lines passing through them and points in the direction of decreasing potential. V A = V B V C From the figures, then the work done to bring a test charge from B to A is given by W BA q q 0 o o V AB V A V B No work is done in moving a charge along the same equipotential surface.

Example 10 Figure below shows a point A at distance 10 m from the positive point charge, q = 5C. Calculate the electric potential at point A and describe the meaning of the answer. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 10 Solution

Example 11 Two point charges, q 1 = +0.3 C and q 2 = 0.4 C are separated by a distance of 6 m as shown in figure below. Calculate the electric potential at point A if point A is at the midpoint of q 1 and q 2. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 11 Solution

Example 12 Two point charges, q 1 = +12 nc and q 2 = 12 nc are separated by a distance of 8 cm as shown in figure below. Determine the electric potential at point P. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 12 Solution

Potential Difference V V final V initial W q 0 OR V AB V A V B W q BA o

Potential Difference The work done to bring a charge from one point to another in the field does not depend on the path taken (because the work done by conservative force).

Example 13 Two points, S and T are located around a point charge of +5.4 nc as shown in figure below. Calculate a) the electric potential difference between points S and T. b) the work done in bringing a charge of 1.5 nc from point T to point S. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 13 Solution

Changes in Potential Energy From the definition of electric potential difference, V ΔV W q 0 and W U ΔV U q 0 Therefore the change in a potential energy is given by U q 0 ΔV U final U initial V final V initial

Potential Energy of A System The total electric potential energy of the system of charges is the total work done to bring all the charges from infinity to their final positions. The total potential energy, U can be expressed as U k q1q r 12 2 q1q r 13 3 q2q r 23 3

Extra Note In the system of charges, suppose there were originally no charges at the points A, B and C as in the figure above. Step 1: The charge q 1 is brought from infinity and placed at point A. Since originally there were no charges, the charge q 1 does not experience any electric force when it is brought from infinity, that is F = 0. Hence V A = 0. Since U 1 = q 1 V A, hence U1 0

Extra Note Step 2: With the charge q 1 fixed at point A, an electric field is produced by the q 1. Electric potential to bring the charge q 2 from infinity to the point B at a distance of r 12 from A is V Since U 2 = q 2 V B, hence U B 2 kq r 12 1 kq q r 1 2 12

Extra Note With the charge q 1 fixed at point A and q 2 at the point B, the electric potential at the point C is V C kq r kq r 1 2 13 23 Since U 3 = q 3 V C, hence U 3 kq q r kq q 1 3 2 3 r 13 23

Extra Note Therefore, the electric potential energy of the system, U U U 0 1 U2 U3 kq1q r k kq1q r 12 12 2 q1q r 12 2 2 kq kq1q r 13 q1q r 13 3 3 3 q r 1 13 kq r 23 q r 2 q2q r 23 3 2 23 q 3

Example 14 Two point charges, Q 1 = +2.0 C and Q 2 = 6.0 C, are placed 4.0 cm and 5.0c m from a point P respectively as shown in figure below. a. Calculate the electric potential at P due to the charges. b. If a charge Q 3 = +3.0 C moves from infinity to P, determine the change in electric potential energy for this charge. c. When the charge Q 3 at point P, calculate the electric potential energy for the system of charges. (Given Coulomb s constant, k = 9.0 10 9 N m 2 C -2 )

Example 14 Solution

1.4 Charge In A Uniform Electric Field Explain quantitatively with the aid of a diagram the motion of charge in a uniform electric field. Use E V d for uniform electric field. Cases: Stationary charge Charge moving perpendicularly to the field Charge moving parallel to the field Charge in dynamic equilibrium Learning Objectives

Uniform Electric Field A uniform electric field is represented by a set of electric field lines which are straight, parallel to each other and equally spaced It can be produced by two flat parallel metal plates which is charged, one with positive and one is negative and is separated by a distance. Direction of E: (+) ve plat to ( ) plat

Case 1 : Stationary Charge Positive stationary charge Negative stationary charge Force experienced by charge is in the same direction as electric field, E. Force experienced by charge is in the opposite direction as electric field, E.

Case 1 : Stationary Charge Consider a stationary particle of charge q o and mass m is placed in a uniform electric field E, the electric force F e exerted on the charge is given by F e q o E Since only electric force exerted on the particle, thus this force contributes the nett force, F and causes the particle to accelerate. According to Newton s second law, then the magnitude of the acceleration of the particle is F q F o e E ma ma a q e m E

Case 1 : Stationary Charge Because the electric field is uniform (constant in magnitude and direction) then the acceleration of the particle is constant. If the particle has a positive charge, its acceleration is in the direction of the electric field. If the particle has a negative charge (electron), its acceleration is in the direction opposite the electric field.

Case 2 : Charge moving perpendicularly to the field Positive charge Negative charge The positive charge will be deflected and moves along a parabolic path towards the negative plate. The positive charge moves under the influence of the electric force which is at the same direction as electric field lines. The negative charge will be deflected and moves along a parabolic path towards the positive plate. The negative charge moves under the influence of the electric force which is opposite direction to the electric field lines.

Case 2 : Charge moving perpendicularly to the field Consider an electron (e) with mass, m e enters a uniform electric field, E perpendicularly with an initial velocity u, the upward electric force will cause the electron to move along a parabolic path towards the upper plate. The electric force F e exerted on the charge, F e q o E

Case 2 : Charge moving perpendicularly to the field From Newton s second law, F ma Therefore the magnitude of the electron s acceleration is given by ee ay (direction: upward) a 0 m ; x e F q F o e E ma ma

u x - component u y - component 0 v v x u v y u y a y t v 2 y u 2 y 2a y s y The path makes by the electron is similar to the motion of a ball projected horizontally above the ground s a s x 0 u x t s s y y u 1 2 y t u y a y 1 2 ayt 2 v ee m e y t

Case 3 : Charge moving parallel to the field F e and v in the same direction F e and v in the opposite direction The electric force on the positive charge is in the same direction as to its motion. The positive charge accelerates along a straight line. The electric force on the positive charge is in the opposite direction to its motion. The positive charge decelerates along a straight line.

Case 4 : Charge in dynamic equilibrium Between electric force and weight Particle weight is at the opposite direction to the electric force. F e W Dynamic equilibrium means the charge moves with constant velocity. Only particles with this constant speed can pass through without being deflected by the fields.

Uniform Electric Field The graph is a straight line with negative constant gradient, thus E V r 0 V d 0 E V d

Example 15 Two parallel plates are separated 5.0 mm apart. The electric field strength between the plates is 1.0 10 4 N C 1. A small charge of +4.0 nc is moved from one conducting plate to another. Calculate a. the potential difference between the plates b. the work done on the charge.

Example 15 Solution

Example 16 If the plates are horizontal and separated by 1.0 cm, the plates are connected to 100 V battery, the magnitude of the electric field is 1.0 10 4 N C -1. If an electron is released from the rest at the upper plate, determine a. the acceleration of the electron. b. the speed and the kinetic energy required to travel to the lower plate. c. the time required to travel to the lower plate. [mass of electron = 9.11 10-31 kg; charge of electron = 1.60 10-19 C ]

Example 16 Solution

Example 17 The figure shows a section of the deflection system of a cathode ray oscilloscope. An electron travelling at a speed of 1.5 10 7 m s -1 enters the space between two parallel metal plates 60 mm long. The electric field between the plates is 4.0 10 3 V m -1. a. Copy the figure, sketch the path of the electron in between plates, and after emerging from the space between the plates. b. Find the acceleration of the electron between the plates. c. Determine the velocity when it emerges from the space between the plates.

Example 17 Solution

Summary Electrostatic Coulomb s Law F Qq 2 r Electric Field E F q Electric Potential V W q o Electric Field For Point Charge E kq 2 r Charge in a Uniform Electric Field E V d Equipotential Surface Electric Potential Energy U kqq 2 r