Transcript
ELECTRIC FORCES AND FIELDS
Electric charge is produced when electrons are separated from neutral atoms so that one object has more electrons than protons and another has more protons than electrons.
The charge on the electron is negative so an excess of electrons gives an object a negative charge.
An object that has lost electrons so that it has fewer electrons than protons is considered to have a positive charge.
The basic unit of charge in the metric system is the Coulomb and corresponds to the amount of charge associated with 6.25 x 1018 electrons or protons.
The charge on one electron is -e = - 1.6 x 10-19 C.
The charge on one proton is +e = + 1.6 x 10-19 C.
Above is an illustration of the positive nucleus surrounded by the negative electrons. If the number of electrons is the same as the number of protons the atom is said to be neutral.
Example
A metal plate has a charge of - 3.0 ?C and a rod has a charge of + 2.0 ?C. How many electrons must be transferred from the plate to the rod so that they have the same charge?
1 Coulomb is 6.25 x 1018 electrons.
The law of conservation of electric charge says that during any process, the net electric charge of an isolated system remains constant. Chemical reactions, electric circuits, and radioactive decay provide examples of this law.
Fundamental characteristic of electric charges says like charges repel and unlike charges attract.
Electric conductors and insulators differ much in the same way heat conductors and insulators differ.
If a substance has electrons that are not tightly bound in molecules, these relatively free electrons can transmit electric energy as well as heat energy. In substances where these free electrons are not present, neither electric or heat energy is transmitted as well.
Two methods of placing an electric charge on an object are conduction and induction.
In charging by conduction, a charged object touches an uncharged object. Charge flows from the charged object to the uncharged object and they are both left with the same type of charge.
In charging by induction, the charged object is brought near the uncharged object causing electrons to be driven to or drawn from the ground. Once the grounding connection is removed and the original charged object is removed, a charge remains on the originally uncharged object opposite in sign to the original charge.
Coulomb's Law
Coulomb's law states that the electric force that exists between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The equation is:
F = (kq1q2)/r2
where q1 and q2 are the two charges expressed in Coulombs, r is the distance between them expressed in meters and k is Coulomb's constant with a value of 8.99 x 109 Nm2/C2 in empty space. The direction of the electrostatic force is along the line connecting the two charges. It is repulsive if they are alike and attractive if they are opposite.
Example
Two tiny spheres have the same mass of 2.0 x 10-6 kg and carry charges of equal magnitude. The gravitational force each exerts on the other is balanced by the electric force. Are the charges both positive, both negative, or one positive and one negative? Find the magnitude of the two charges. G = 6.67 x 10-11Nm2/kg2.
The permittivity of free space is another way to express Coulomb's constant and is ?0 = 1/4?k or 8.85 x 10-12 C2/Nm2. This will be handy when we look at electric flux.
When more than two point charges are involved, we must find the vector sum of all of the electric forces acting on the point charge of interest.
If they are arranged in a linear fashion, we have a one dimensional problem and the forces can be in the positive direction or the negative direction
If the charges are arranged in a plane so that they do not form a line, we must consider X and Y components or use trigonometry to solve parts of a triangle.
Example
Three point charges are fixed in place as in the diagram. q1 has a charge of +8.00 ?C, q2 has a charge of -5.00 ?C and q3 has a charge of +5.00 ?C. (a) Determine the net force exerted on q1 by the other two charges and (b) if q1 had a mass of 1.50 g and it were free to move, what would be its acceleration?
An electric field is defined as a region of space in which an electric force will be experienced by a charged particle that is placed there. The SI unit of electric field strength is the newton per coulomb(N/C). The formula used to calculate electric field strength is:
? = F/q
Where ? is the electric field strength, F is the force experienced by the charge at that location, and q is the magnitude of the charge placed in the field. Electric fields are vectors and must be treated as such in calculation.
Example
A tiny ball with a mass of 0.012 kg carries a charge of -18.0 ?C. What magnitude and direction electric field is needed to cause the ball to float above the ground?(g = 9.8 m/s2)
The electric field produced by a point charge is symmetrical around the location of the point with field lines radiating outward from a positive charge and radiating inward towards a negative charge.
The magnitude of the electric field strength produced by a point charge varies inversely as the square of the distance from the charge's location. The formula is:
? = kq/r2
where q is the magnitude of the point charge causing the field to exist. This equation indicates that there will be shells surrounding the point charge on which the electric force will not change. Movement from one shell to another will change the electric force and require a change in energy. Movement on a shell will change neither the electric force or electric potential energy.
Example
A 3.0 ?C point charge is placed in an external uniform electric field with magnitude 1.6 x 104 N/C. Where relative to the point charge is the electric field zero?
The case of two point charges relatively near each other and with opposite charges of the same magnitude is called an electric dipole. The product of the magnitude of one of the charges and the distance between them is called the electric dipole moment. Polar molecules such as H2O and HCl have dipole moments.
Electric dipoles produce curved electric field lines. The electric field vector direction is tangent to the electric field line at any point.
The parallel plate capacitor is a device which consists of two metal plates arranged so that the relatively large, flat sides face each other. The electric field strength within the capacitor can be calculated using the formula:
? = q/?0A = ?/ ?0
where A is the area of one plate and ? is the charge density on the plate. Near the edges of the plates, the field strength will be different, but over most of the volume between the plates this equation can be used to find the field strength.
Example
A small plastic ball with a mass of 6.50 x 10-3kg and a charge of +0.150 ?C hangs in equilibrium between the plates of a capacitor as in the diagram. If the area of each plate is 0.0150 m2, find the magnitude of the charge on each plate.
Shielding and the electric field inside a conductor
If a net charge exists within a solid conductor, all of the individual charges would repel each other. This net repulsive force would cause them to move apart until they reached the outer boundary of the conductor. This gives us the statement under equilibrium conditions, all charge resides on the outside of a conductor.
The charges arrange themselves so that the electric field within the conductor is zero.
This is the basis for electric shielding. Any charge inside the conductor will experience no electric force due to any electric field outside the conductor.
Gauss's Law
Consider a positive point charge surrounded by a spherical surface of radius r.
This closed sphere is called a Gaussian surface
Since the equation for the electric field strength at this surface is:
? = kq/r2
and can be rewritten as:
? = q/?04?r2
since k = 1/4??o. Also A = 4?r2.
Then
? = q/?0A
and
?A= q/?0
The product ?A is called electric flux and is represented by the symbol ?. So the equation becomes:
? = ?A= q/?0
If the surface is not a sphere, we must find the component of the electric field that is perpendicular to the surface since the electric flux depends only on the component of the electric field that passes through the surface at right angles to the surface. A field component parallel to the surface does not actually pass through it.
The component of the electric field passing through the surface is ?cos? where ? is the angle between the direction of the electric field and the normal to the surface.
The equation for electric flux becomes:
? = ?cos?A= q/?0
for a flat surface.
If the surface is not spherical or flat, we must add the flux through all of the individual surfaces to get the total flux. The equation becomes:
? = ??cos??A= q/?0
It can be shown that this equation is true for any shape surface. The SI unit of electric flux is the Nm2/C.
Example
Two planar surfaces intersect and are perpendicular. Surface 1 has an area of 1.7 m2 and surface 2 has an area of 3.2 m2. The electric field is uniform and has a magnitude of 250 N/C. Find the electric flux through (a) surface 1, (b) surface 2, and (c) the total electric flux through both surfaces.
