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PHYS 1008 ELEMENTARY UNIVERSITY PHYSICS II Instructor: Dr. Wafia Bensalem Textbook & Reference: Physics, Giambattista, Richardson, Richardson, McGraw Hill, 2016.CHAPTER 1
Electric Forces and Fields
4. THE ELECTRIC FIELD If a point charge q is in the vicinity of other charges, it experiences an electric force
.
The electric field (symbol ) at any point is defined to be the electric force per unit charge at that point. NB:
is a vector quantity
Once we know the electric field at some point it is easy to calculate the electric force
on any point charge q placed there:
Example 16.5 A small sphere of mass 5.10 g is
hanging vertically from an insulating thread that is 12.0 cm long. By charging some nearby flat metal plates, the sphere is subjected to a horizontal electric field of magnitude 7.20 × 105 N/C. As a result, the sphere is displaced 6.00 cm horizontally in the direction of the electric field. (a) What is the angle that the thread makes with the vertical? (b) What is the tension in the thread? (c) What is the charge on the sphere?Slide 5 Strategy • The electric force on the sphere is given by . The figure shows that the sphere is pushed to the right by the field; therefore,
is to the right.
• Since and
have the same direction, the charge on the sphere is positive.
• After drawing an FBD showing all the forces acting on the sphere, we set the net force on the sphere equal to zero since it hangs in equilibrium.Solution
(a)
Solution
(b)
Solution (c)
Electric Field due to a Point Charge What is the electric field [due to a single point charge Q] at a position P, when P is at a distance r from Q
Let q be a positive test charge placed at P. Coulomb’s law says that the force acting on the test charge has a magnitude:
The electric field strength at P is then:Electric field strength at P Source charge
Coulomb’s constant
Distance between P and the source charge ❖ Principle of Superposition The electric field at any point is the vector sum of the field vectors at that point caused by each charge separately.
QUIZ 4 True or false? A charge of +3µC is at a point P where the electric field is directed to the right and has a magnitude of 4 x 106 N/C. If the charge is replaced with a -3µC charge, electric field at P will be directed to the left.ANSWER False. Nothing happens to the electric field. It is created not by the charge put on the point P, but by the source charge.Electric Field Lines Electric field lines are a set of continuous lines that represent the electric field vector in space. • The direction of the electric
field vector at any point is tangent to the field line passing through that point and in the direction indicated
by arrows on the field line. • The electric field is strong where field lines are close together and weak where they are far apart.Rules for Sketching Field Lines • Electric field lines can start only on positive charges and can end only on negative charges. • The number of lines starting on a positive charge (or ending on a negative charge) is proportional to the magnitude of the charge.
(The total number of lines you draw is arbitrary; the more lines you draw, the better the representation of the field.)• Field lines never cross. The electric field at any point has a unique direction; if field lines crossed, the field would have two directions at the same point.
How to observe them? Small pieces of thread on an oil surface align with the electric field.Field Lines for a Point Charge • The electric field due to a point charge is radial. It goes away from a positive charge or toward a negative charge.
• The lines are close together near the point charge, where the field is strong, and are more spread out farther from the point charge, showing that the field strength diminishes with distance.
QUIZ 5If E1 is the magnitude of the electric field at the point P due to Q1 and E2 is the magnitude of the electric field at the point P due to Q2; the net electric field at the point P is: a) E1+ E2 b)22 2 E1+ E c)E i E j 1+ 2 d)E j E i 1− 2 Q2(-) P X Q1(+) ANSWER D. The net electric field is the vector sum of the two electric fields due to Q1 And Q2. Adding vectors does NOT mean adding magnitudes! The electric field is a vector (NOT a scalar) quantity!Electric Field due to a Dipole A pair of point charges with equal and opposite charges that are near one another is called a dipole (literally two poles ). Sketching some field lines gives an approximate idea of the electric field.
QUIZ 6
In the figure below, the relation between the magnitudes of the electric fields at points A, B, and C ( EA, EB and EC ) is: A) EC < EB < EA B) EB < EA < EC C) EA < EB < EC D) EC < EA < EBANSWER A. The field is greatest at point A because this is where the field lines are closest together. The absence of lines at point C indicates that the electric field there is zero.Example 16.6 Two point charges are located on the x -axis. Charge q1 = +0.60 μC is located at x = 0; charge q2 = − 0.50 μC is located at x = 0.40 m. Point P is located at x = 1.20 m. What is the magnitude and direction of the electric field at point P due to the two charges?
Strategy • We can determine the field at P due to q1and the field at P due to q2separately using Coulomb’s law and the definition of the electric field. • In each case, the electric field points in the direction of the electric force on a positive test charge at point P . • The sum of these two fields is the electric field at P .Solution
Solution
Solution
Example 16.7 Three point charges are placed at the corners of a rectangle, as shown in the figure. (a) What is the electric field due to these three charges at the fourth corner, point P? (b) What is the acceleration of an electron located at point P? Assume that no forces other than that due to the electric field act on it.
Strategy (a) After determining the magnitude and direction of the electric field at point P due to each point charge individually, we use the principle of superposition to add them as vectors. (b) Since we have already calculated
at point P, the force on the electron is , where q = −e is the charge of the electron.
Solution (a)
Solution
(a)
Solution (b) The force on the electron is . Its acceleration is then . The electron charge qe = −e and mass meare given in Table 16.1. The acceleration has magnitude a = eE/me = 6.2 × 1016 m/s2. The direction of the acceleration is the direction of the electric force, which is opposite the direction of
since the electron’s charge is negative.
Example 16.8 A thin metallic spherical shell of radius R carries a total charge Q , which is positive. The charge is spread out evenly over the shell’s outside surface. Sketch the electric field lines in two different views of the situation: (a) the spherical shell is tiny, and you are looking at it from distant points; (b) you are looking at the field inside the shell’s cavity. In (a), also sketch field vectors at two different points outside the shell.
Strategy • Since the charge on the shell is positive, field lines begin on the shell. • A sphere is a highly symmetrical shape: standing at the center, it looks the same in any chosen direction. This symmetry helps in sketching the field lines.Solution (a)
Solution (b)
The lines must start on the shell and point radially toward the center. They would cross. So there can be NO field lines (No electric field) inside the shell.Application of Electric Fields: Electrolocation The Gymnarchus niloticus (Nile River)
navigates With great precision although it is nearly blind. The Gymnarchus
generates an electric field resembling that of a dipole. Slight changes in the Electric field are interpreted as the presence of nearby objectsApplication : Lightning Rods
The lightning-rod is an excellent conductor. The current flows to ground without causing any heat damage in the building.Slide 39