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One coulomb represents the negative of trons. For comparison, a copper cube I cin on a side contains it electrons. About 10" electrons pass through the glowing hlament of a flashlight bulb every second. In electrostatics problems (that is, problems that involve charges at rest), it's very unusual to encounter charges as large as I coulomb. Two I-C charges sepa rated by I'm would exert forces on each other of magnitude 9 % 10" N (about 1 million tons)! The total charge of all the electrons in a copper one-cent coin is even greater, about 1.4 X 10' C, which shows that we can't disturb electric neu trality very much without using enormous forces. More typical values of charge range from about 10 " to about 10 " C. The microcoulomb ( 1 AC = 10 'C) and the nanocoulomb ( I nc = 10 ' C ) are often used as practical units of charge Electric force versus gravitational force An a particle ("alpha") is the nucleus of a helium atom. It has EXECUTE: The ratio of the electric force to the gravitational mass m = 6.64 X 10 27kg and charge q = + 2e = 3.2 x 10 "C. force is Compare the force of the electric repulsion between two a F. 9.0 X 10' N . m / C2 ( 3.2 x 10 " C ) ? particles with the force of gravitational attraction between them. ATE, Gm? 6.67 X 10 " N . m/kg (6.64 x 10 2 kg)? SOLUTION = 3.1 X 1035 IDENTIFY: This problem involves Newton's law for the gravita- EVALUATE: This astonishingly large number shows that the gravi- tional force F, between particles (see Section 12.1) and Coulomb's tational force in this situation is completely negligible in compari law for the electric force F. between point charges. We are asked to son to the electric force. This is always true for interactions of compare these forces. so our target variable is the ratio of these atomic and subatomic particles. (Notice that this result doesn't two forces. Fe Fe depend on the distance r between the two a particles.) But within objects the size of a person or a planet, the positive and negative SET UP: Figure 21.11 shows our sketch. The magnitude of the charges are nearly equal in magnitude, and the net electric force is repulsive electric force is given by Eq. (21.2): usually much smaller than the gravitational force. ( F. = - 21.11 Our sketch for this problem. ATTER 7 2 q = 3.2X 10-19 C The magnitude F of the attractive gravitational force is given by m = 6.64X 10-27 kg Eq. (12.1)