Part 3 The electric field can push AND pull charges Because there are two types of charges (+ and -) The gravitational field can only pull Only positive masses Charges cause an electrical field. Each charge creates its own electric field in the surrounding space. Electric Field is a vector quantity Has magnitude and direction Use Electric Field lines(lines of force) to show the Electric Field Electric field direction from charge that creates the field: Away from positive charge Toward negative charges The lines begin on positive charges and end on negative charges. Draw lines perpendicular to charge Lines never cross Electric field strength: The closer to the charge, the stronger the field The closer the field lines, the stronger the field The farther away, the weaker the field Electric field strength: The number of lines are proportional to strength The more lines the stronger the field Electric field strength: (quantitative) E= 2 kq/d ▪ Depends on Charge and distance E = Electric field strength (Newton/Coulomb, N/C) k = 9.0 x 109 Nm2/C q= charge causing the field (Coulomb, C) d or r = distance (meters, m) A fly accumulates 3.0 x 10-10 C of positive charge as it flies through the air. What is the magnitude and direction of the electric field at a location 2.0 cm away from the fly? q = 3.0 x 10-10C d or r = 2.0 cm = 0.020 _____ m E=? equation E = kq/ r2 E = (9.0 x 109)(3.0x10-10) (0.020)2 E = 6750 N/C Electric Field = force exerted on positive charge If +q = force in same direction (pull) of the field If -q = force in opposite direction (push) of the field The strength of the electric field… E = F/q E = strength of the field (Newton/Coulomb, N/C) F = Force exerted on charge in field (Newton, N) q = charge (Coulomb, C) that sits in the field; experiences the force A charge of -2.0 x10-6C is placed in a uniform electric field of strength 5000 N/C that points downward. What is the magnitude and direction of the force experienced by this charge? A charge of -2.0 x10-6C is placed in a uniform electric field of strength 5000 N/C that points downward. What is the magnitude and direction of the force experienced by this charge? E=F/q 5000 N/C= F / (-2.0 x 10-6C) (-2x10-6 )(5000)=F -0.01N=F magnitude of the force direction? - 2 charge, the force on the electron is opposite the field!