Cycle III Regents Physics

John Dewey High School

Mr. Klimetz

The electric field intensity **E** at a particular point is given by the relationship

where **F****e** is the force (N) exerted on a test charge q (C). The unit for electric field intensity is N/C.

If the electric field intensity at a particular point is known, the force exerted on any amount of charge at that point is

At any point within a charged hollow spherical body, the field intensity is zero. The net force on a test charge, found by adding the forces exerted on all the charges on the hollow sphere, is zero. Therefore, there is no electric field inside the hollow sphere. Outside the hollow sphere, the field in the same as though all the charges were concentrated at the center of the hollow sphere.

The field created by two equal positive point charges is the same in appearance as that created by two equal negative point charges except in the orientation of the arrows: outward from the positive point charges and inward towards the negative point charges. The field lines in the space directly between the pair of identical point charges are deflected laterally and never mutually cross.

The field created by two equal but opposite point charges appears as a single straight and a set of curved lines extending radially outward from the positive point charge and ending radially inward at the negative point charge.

The electric field lines between two parallel but oppositely charged plates generally appear as parallel lines extending perpendicularly from the positive plate and ending at the negative plate. In the case where the distance between the plates is small compared to their size and the charge of the plates is uniformly distributed, the field lines are parallel to each other and perpendicular to the plates (except near the edges where they bulge slightly outward). If a positive test charge is released between the plates, it will be pushed in a straight line from the positive plate to the negative plate. Since the field between the plates is uniformly intense, the magnitude of the force exerted on a test charge is the same everywhere in the region between the plates (except near the edges). Outside the plates, the net force is practically zero, so no fieldlines appear there.

The following features are common to all electric field diagrams:

a. Field lines always begin on the surfaces of positively charged objects and terminate on the surfaces of negatively charged objects.

b. Field lines never intersect one another.

c. Where field lines meet a charged object they are oriented perpendicularly to the surface of the charged object.

d. Field lines do not extend below the surface of a charged object.

If an electric field does work on any charged object, such as when the field created by one positive charge repels another positive charge, electric potential energy decreases. Similarly, electric potential energy decreases when the field created by a positive charge attracts a negative charge, pulling the charges closer together. This is analogous to a falling object on Earth. When the gravitational field does work, gravitational potential energy does work, and gravitational potential energy decreases. In accordance with the Law of Conservation of Energy, the lost potential energy is converted into an equal amount of energy of another type, such as kinetic energy or heat.

If work is done against an electric field, such as when a positive test charge is moved closer to another positive charge, electric potential energy increases. Similarly, electric potential energy increases when a negative charge is moved away from a positive charge. This is analogous to the lifting of an object on Earth. When work is done against the gravitational field, gravitational potential energy increases. Once again, the gain in potential energy must correspond to a loss of energy of some other type, in order that energy be conserved. The work done by (or against) an electric field in moving a charge from one point to another is independent of the path taken. In other words, the electric force is a conservative force, as is gravity.

At a point in a field created by a negative charge, the force is attractive. Thus, the work done in moving the positive test charge from infinity to the point is negative. (The field does work on the charge.) The closer the point is to the negative charge that initiated the field, the smaller the electric potential.

In summary, electric potential at any given point in an electric field is positive if work must be done against the field to move a positive test charge from infinity to that point, and negative if work is done by the the field to move a positive test charge from infinity to that point.

From the amount of work W, in joules, required to move a charge q, in coulombs, between two points in an electric field, the potential difference V between those two points can be obtained by using the equation\

V = W/q

Thus, the potential difference is expressed in joules per coulomb (J/C). One joule per coulomb is more commonly referred to as one volt. Since the potential difference between two points is measured in volts, it is also known as the voltage.

If it takes 6 J of work to push 3 C of charge from one point to another in an electric field, the potential difference between the two points is 6 J/3 C, or 2 V. If a manufacturer labels a battery 9 V, it means that the electric field created by the two charged terminals will do 9 joules of work on every coulomb that is pushed by the battery from one terminal to the other. Since the quantity of work done is independent of the path taken between the terminals, that characteristics of the materials and appliances connected to the battery play no role in determining the voltage.

If the voltage between two points is known, the work done in moving any given amount of charge between those points can be found by the equation

W = qV

V = **E**d

where d is the distance in meters between the two points.

Solving for **E** we obtain

Thus, a uniform electric field intensity can be expressed in volts per meter (V/m). Since **E** was defined earlier in units of newtons per coulomb (N/C), we can conclude that 1 V/m is equivalent to 1 N/C.

1. Calculate the charge possessed by two objects q1 and q2 with values of -12 C and +5 C after they are connected with a conducting wire.

2. Calculate the **F**e between two charges each possessing a value of -5 C and separated by a distance of 5 m.

3. Calculate the **E** at a distance of 2 m when a test charge of +3.20 x 10exp-15 C is placed in the field produced by a charge of +1.60 C.

4. Calculate the amount of work done in moving +5 C across a potential difference of 440 volts.

5. Calculate the number of electrons which pass a given point in a wire which is experiencing a current of 20 amperes over a 1 hour time period.

6. Assuming a uniform, steady state flow of current in a wire which only permits one electron to exit at a time, calculate the amount of time it takes between successive electrons to exit the wire if it is experiencing a current of 6 amperes.