Properties of Electric Charge Electric Charge There are two kinds of electric charge. like charges repel unlike charges attract Electric charge is conserved. Positively charged particles are called protons. Uncharged particles are called neutrons. Negatively charged particles are called electrons. Properties of Electric Charge, continued The Milikan Experiment Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge. Charge is measured in coulombs (C). The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton. e = 1.60 176 x 10 19 C Milikan s Oil Drop Experiment Transfer of Electric Charge An electrical conductor is a material in which charges can move freely. An electrical insulator is a material in which charges cannot move freely.
Transfer of Electric Charge, continued Charging by Induction Insulators and conductors can be charged by contact. Conductors can be charged by induction. Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor. Transfer of Electric Charge, continued A surface charge can be induced on insulators by polarization. With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation. Coulomb s Law Two charges near one another exert a force on one another called the electric force. Coulomb s law states that the electric force is proportional to the magnitude of each charge and inversely proportional to the square of the distance between them. q1q electric = kc r ( charge 1)( charge ) electric force = Coulomb constant ( distance) Coulomb s Law, continued Superposition Principle The resultant force on a charge is the vector sum of the individual forces on that charge. Adding forces this way is an example of the principle of superposition. When a body is in equilibrium, the net external force acting on that body is zero.
Sample Problem Consider three point charges at the corners of a triangle, as shown at right, where q 1 = 6.00 10 C, q =.00 10 C, and q 3 =.00 10 C. ind the magnitude and direction of the resultant force on q 3. 1. Define the problem, and identify the known variables. Given: q 1 = +6.00 10 C r,1 = 3.00 m q =.00 10 C r 3, = 4.00 m q 3 = +.00 10 C r 3,1 =.00 m θ = 37.0º Unknown: 3, =? Diagram: Tip: According to the superposition principle, the resultant force on the charge q 3 is the vector sum of the forces exerted by q 1 and q on q 3. irst, find the force exerted on q 3 by each, and then add these two forces together vectorially to get the resultant force on q 3.. Determine the direction of the forces by analyzing the charges. The force 3,1 is repulsive because q 1 and q 3 have the same sign. The force 3, is attractive because q and q 3 have opposite signs. 3. Calculate the magnitudes of the forces with Coulomb s law. (.00 10 C)( 6.00 10 C) ( ) = kc = 8.99 10 3 1 9 3,1 3,1 8 1.08 10 N (.00 10 C)(.00 10 C) ( ) = kc = 8.99 10 3 9 3, 3,1 q q N m ( r3,1) C.00 m = q q N m ( r3,) C 4.00m.6 10 N = 4. ind the x and y components of each force. At this point, the direction each component must be taken into account. 3,1 : x = ( 3,1 )(cos 37.0º) = (1.08 10 8 N)(cos 37.0º) x = 8.63 10 N y = ( 3,1 )(sin 37.0º) = (1.08 10 8 N)(sin 37.0º) y = 6.0 10 N 3, : x = 3, =.6 10 N y = 0 N. Calculate the magnitude of the al force acting in both directions. x, = 8.63 10 N.6 10 N = 3.01 10 N y, = 6.0 10 N + 0 N = 6.0 10 N
6. Use the Pythagorean theorem to find the magnitude of the resultant force. = ( ) + ( ) = (3.01 10 N) + (6.0 10 N) 9 9 3, x, y, 3, = 7.16 10 N 7. Use a suitable trigonometric function to find the direction of the resultant force. In this case, you can use the inverse tangent function: y, 6.0 10 N tanϕ = = 3.01 10 N x, ϕ = 6.º Coulomb s Law, continued The Coulomb force is a field force. A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects. An electric field is a region where an electric force on a test charge can be detected. The SI units of the electric field, E, are newtons per coulomb (N/C). The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge. Electric ields and Test Charges, continued Electric field strength depends on charge and distance. An electric field exists in the region around a charged object. Due to a Point Charge q E = kc r charge producing the field electric field strength = Coulomb constant ( distance)
Calculating Net Electric ield Sample Problem A charge q 1 = +7.00 µc is at the origin, and a charge q =.00 µc is on the x- axis 0.300 m from the origin, as shown at right. ind the electric field strength at point P,which is on the y-axis 0.400 m from the origin. 1. Define the problem, and identify the known variables. Given: q 1 = +7.00 µc = 7.00 10 6 C r 1 = 0.400 m q =.00 µc =.00 10 6 C r = 0.00 m θ = 3.1º Unknown: E at P (y = 0.400 m) Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge. 6 q 9 1 7.00 10 C E1 = kc = ( 8.99 10 N m /C ) 3.93 10 N/C r = 1 (0.400 m) E q.00 10 C 8.99 10 N m /C 1.80 10 N/C ( ) 6 = kc = r (0.00 m) = 9 3. Analyze the signs of the charges. The field vector E 1 at P due to q 1 is directed vertically upward, as shown in the figure, because q 1 is positive. Likewise, the field vector E at P due to q is directed toward q because q is negative. 4. ind the x and y components of each electric field vector. or E 1 : E x,1 = 0 N/C E y,1 = 3.93 10 N/C or E : E x, = (1.80 10 N/C)(cos 3.1º) = 1.08 10 N/C E y,1 = (1.80 10 N/C)(sin 3.1º)= 1.44 10 N/C
. Calculate the al electric field strength in both directions. E x, = E x,1 + E x, = 0 N/C + 1.08 10 N/C = 1.08 10 N/C E y, = E y,1 + E y, = 3.93 10 N/C 1.44 10 N/C =.49 10 N/C 6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector. (, ) (, ) E = E + E E E x y ( 1.08 10 N/C) (.49 10 N/C) = + =.71 10 N/C 7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector. In this case, you can use the inverse tangent function: Ey,.49 10 N/C tanϕ = = E 1.08 10 N/C x, ϕ = 66.0 8. Evaluate your answer. The electric field at point P is pointing away from the charge q 1, as expected, because q 1 is a positive charge and is larger than the negative charge q. Electric ield Lines Rules for Drawing Electric ield Lines The number of electric field lines is proportional to the electric field strength. Electric field lines are tangent to the electric field vector at any point.
Rules for Sketching ields Created by Several Charges Conductors in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor. Any excess charge on an isolated conductor resides entirely on the conductor s outer surface. The electric field just outside a charged conductor is perpendicular to the conductor s surface. On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.