 # AP Physics Summer Institute 2007. ELECTROSTATICS.

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AP Physics Summer Institute 2007

ELECTROSTATICS

ELECTROSTATICS: The study of the behavior of stationary charges ELECTRIC CHARGE There are two types of electric charge, arbitrarily called positive and negative. Rubbing certain electrically neutral objects together (e.g., a glass rod and a silk cloth) tends to cause the electric charges to separate. In the case of the glass and silk, the glass rod loses negative charge and becomes positively charged while the silk cloth gains negative charge and therefore becomes negatively charged. After separation, the negative charges and positive charges are found to attract one another.

If the glass rod is suspended from a string and a second positively charged glass rod is brought near, a force of electrical repulsion results. Negatively charged objects also exert a repulsive force on one another. These results can be summarized as follows: unlike charges attract and like charges repel.

CONSERVATION OF ELECTRIC CHARGE In the process of rubbing two solid objects together, electrical charges are not created. Instead, both objects contain both positive and negative charges. During the rubbing process, the negative charge is transferred from one object to the other leaving one object with an excess of positive charge and the other with an excess of negative charge. The quantity of excess charge on each object is exactly the same.

The law of conservation of electric charge: "The net amount of electric charge produced in any process is zero." Another way of saying this is that in any process electric charge cannot be created or destroyed, however, it can be transferred from one object to another. Charged comb attracts neutral bits of paper. Charged comb attracts neutral water molecules.

The SI unit of charge is the coulomb (C). 1 C = 6.25 x 10 18 electrons or protons The charge carried by the electron is represented by the symbol -e, and the charge carried by the proton is +e. A third particle, which carries no electrical charge, is the neutron. e = 1.6 x 10 -19 C m electron = 9.11 x 10 -31 kg m proton = 1.672 x 10 -27 kg m neutron = 1.675 x 10 -27 kg.

Experiments performed early in this century have led to the conclusion that protons and neutrons are confined to the nucleus of the atom while the electrons exist outside of the nucleus. When solids are rubbed together, it is the electrons that are transferred from one object to the other. The positive charges, which are located in the nucleus, do not move. Rubber scrapes electrons from fur atoms.

INSULATORS AND CONDUCTORS An insulator is a material in which the electrons are tightly held by the nucleus and are not free to move through the material. There is no such thing as a perfect insulator, however examples of good insulators are: glass, rubber, plastic and dry wood. A conductor is a material through which electrons are free to move through the material. Just as in the case of the insulators, there is no perfect conductor. Examples of good conductors include metals, such as silver, copper, gold and mercury.

A few materials, such as silicon, germanium and carbon, are called semiconductors. At ordinary temperature, there are a few free electrons and the material is a poor conductor of electricity. As the temperature rises, electrons break free and move through the material. As a result, the ability of a semiconductor to conduct improves with temperature.

COULOMB’S LAW Coulomb’s Law states that two point charges exert a force (F) on one another that is directly proportional to the product of the magnitudes of the charges (q) and inversely proportional to the square of the distance (r) between their centers. The equation is: F = electrostatic force (N) q = charge (C) k = 9x10 9 N. m 2 /C 2 r = separation between charges (m)

The value of k can also be expressed in terms of the permittivity of free space (ε o ): 9x10 9 N. m 2 /C 2 The proportionality constant (k) can only be used if the medium that separates the charges is a vacuum. If the region between the point charges is not a vacuum then the value of the proportionality constant to be used is determined by dividing k by the dielectric constant (K). For a vacuum K = 1, for distilled water K = 80, and for wax paper K = 2.25

1. Two charges q 1 = - 8 μC and q 2 = +12 μC are placed 120 mm apart in the air. What is the resultant force on a third charge q 3 = - 4 μC placed midway between the other charges? q 1 = - 8 μC q 2 = +12 μC q 3 = - 4 μC r = 0. 120m -q1-q1 +q2+q2 -q3-q3 F1F1 F2F2 FRFR = 80 N = 120 N F R = 80 + 120 = 200 N, to the right

2. Three charges q 1 = +4 nC, q 2 = -6 nC and q 3 = -8 nC are arranged as shown. Find the resultant force on q 3 due to the other two charges. q 1 = +4 nC q 2 = -6 nC q 3 = -8 nC F2F2 F1F1 37˚ θ = 2.88x10 -5 N = 6.75x10 -5 N FRFR

F2F2 F1F1 37˚ θ From the FBD: Σ F x = F 2 - F 1 cos 37˚ = (6.75x10 -5 ) - (2.88x10 -5 )(cos 37˚) = 4.45x10 -5 N Σ F y = F 1 sin 37˚ = (2.88x10 -5 )(sin 37˚) = 1.73x10 -5 N = 4.8x10 -5 N θ = 21˚ F R (4.8x10 -5 N, 21˚) FRFR

ELECTRIC FIELD An electric field is said to exit in a region of space in which an electric charge will experience an electric force. The magnitude of the electric field intensity is given by: Units: N/C

The direction of the electric field intensity at a point in space is the same as the direction in which a positive charge would move if it were placed at that point. The electric field lines or lines of force indicate the direction. The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.

The electric field intensity E at a distance r from a single charge q can be found as follows: Units: N/C

When more than one charge contributes to the field, the resultant field is the vector sum of the contributions from each charge. Units: N/C

3. Two point charges q 1 = -6 nC and q 2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field a. At point A q 1 = -6 nC q 2 = +6 nC E1E1 E2E2 ERER = 3.38x10 4 N/C, left = 8.44x10 3 N/C, left

E1E1 E2E2 ERER E R = E 1 + E 2 = 3.38x10 4 + 8.44x10 3 = 4.22x104 N/C, left

3. Two point charges q 1 = -6 nC and q 2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field b. At point B q 1 = -6 nC q 2 = +6 nC E1E1 37º E2E2 ERER θ = 6.67x10 3 N/C, down = 2.4x10 3 N/C at 37˚

E1E1 37º E2E2 ERER θ From FBD Σ E x = - E 2 cos 37˚ = - (2.4x10 3 )(cos 37˚) = -1916.7 N/C Σ F y = E 2 sin 37˚- E 1 = (2.4x10 3 )(sin 37˚) - (6.67x10 3 ) = - 5225.6 N/C = 5566 N/C

E1E1 37º E2E2 ERER θ = 70˚ 180˚ + 70˚ = 250˚ E R (5566 N/C, 250˚)

ELECTRIC POTENTIAL The electric potential V is defined in terms of the work to be done on a charge to move it against an electric field. The electric potential V is a scalar quantity defined as the potential energy per unit charge. Units: J/C = Volt (V) As discussed in the energy chapter, only differences in potential are measurable.

We find the potential difference between two points A and B to be equal to: In other words the potential difference between two points is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.

Since the electric potential is defined as the potential energy per unit charge, then the change in potential energy of a charge q when moved between two points A and B is: ΔPE = PE B - PE A = qV BA The potential at a point is defined in terms of a positive charge. The potential due to a positive charge is positive, and the potential due to a negative charge is negative.

The potential is the same at equal distances from a spherical charge. The dashed lines in the figure below are called equipotential lines. Note that the lines of equal potential are always perpendicular to the electric field lines.

POTENTIAL AND ELECTRIC FIELD The potential difference between two oppositely charged plates is equal to the product of the field intensity and the plate separation. W = q V BA W = F d = q Ed (d is distance) V BA = E d Units: V BA = Volts (V) E = V/m or N/C

4. The potential difference between two plates 5 mm apart is 10 kV. Determine the electric field intensity between the two plates. r = 5x10 -3 m V = 10x10 3 V = 2x10 6 V/m

ELECTRIC POTENTIAL due to POINT CHARGES The potential V at a point a distance r from a charge Q is equal to the work per unit charge done against electric forces in bringing a positive charge +q from infinity to that point. In other words, the potential at some point A as shown in the figure, is equal to the potential energy per unit charge. A Units: J/C = Volt (V)

The potential in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge. V = V 1 + V 2 + V 3 +........... + V N =  (Q i /r i ) Units: Volts (V)

5. What is the electric potential at the center of a square of sides equal 1 m if the charges placed at the corners are +1 nC, -2 nC, +3 nC and -4 nC (read in a CW direction) +1-2 - 4+3 The center of the square is equidistant from all four charges, a distance r of V = V 1 + V 2 + V 3 + V 4 = - 25 V

ELECTRIC POTENTIAL ENERGY The electric potential energy of a system composed of a charge q and another charge Q separated by a distance r is equal to the work done against the electric forces in moving a charge +q from infinity to that point. Joules (J)

Whenever a positive charge is moved against an electric field, the potential energy increases; whenever a negative charge is moved against an electric field, the potential energy decreases.

6. A charge of + 2 nC is 20 cm away from another charge of + 4 μC. a. What is the potential energy of the system? q = +2 nC Q = + 4 μC r = 20 cm = 36x10 -5 J b. What is the change in potential energy if the 2 nC charge is moved to a distance of 8 cm from the 4 μC charge? r = 8 cm = 90x10 -5 J ΔPE = 90x10 -5 - 36x10 -5 = 54x10 -5 J