Physics net charge homework help

[Mild]

I've been behind in class. It'd really be helpful if anyone could help me. It's hard, but if you know it, just help me, please. Here it is:
1) Calculate the net charge on a substance consisting of (a) 5 x 10^14 electrons and (b) a combination of 7 x 10^13 protons and 4 x 10^13 electrons.

2) Two identical conducting spheres are placed a distance of 0.3 m between their centers. One is given a charge of 12 x 10^-9 C and the other a charge of -18 x 10^-9 C. (a) Find the electrostatic force exerted on one sphere by the other. (b) The spheres are now connected by a conducting wire. After equilibrium has occurred, find the electrostatic force between the two.

3) The electric force on a point charge of 5 x 10^-9 C at some point is 3.8 x 10^-3 N in the positive X direction. What is the magnitude of the electric field at this location?

4) What are the magnitude and direction of the electric field setup by the proton at the position of the electron in the hydrogen atom?

5) The electrons in a particle beam each have a kinetic energy of 1.6 x 10^-17 J. What are the magnitude and direction of the electric field that will stop these electrons in a distance of 10cm?

6) A proton accelerates from rest in a uniform electric field of 640 N/C. At some later time, its speed is 1.20 x 10^6 m/s. (a) Find the acceleration of the proton. (b) How long does it take the proton to reach this velocity? (c)How far has it moved in this time? (d) What is its kinetic energy at this time?

7) (a) How much work is done by a uniform electric field 200 N/C as the charge moves a distance of 2 cm in the field? (b) What's the difference in potential energy between these two points?

8) Show that units N/C and V/m are equivalent.

9) The difference in potential energy between the accelerating plates of a TV set is about 25,000 V. If the distance between these plates is 1.5 cm, find the magnitude of the uniform electric field in this region.

10) (a) What potential difference is needed to stop an electron with an initial speed of 4.2 x 10^5 m/s? (b)How much energy does this require?

11) At what distance from a point charge of 6 microcoulombs would the potential equal 27,000 V?

12) Calculate the speed of (a) an electron that has a kinetic energy of lev, and (b) a proton that has a kinetic energy of 1eV.

13) How much energy is gained by a charge of 75 microcoulombs moving through a potential difference of 90V? Express your answer in (a) joules and (b) electronvolts.

 

[maverick280857]

How about showing your work first?

 

[wais]

Sure, I'd be happy to help you with your physics net charge homework. It's completely understandable to fall behind in class, so don't worry, we'll work through these problems together.

1) To calculate the net charge, we need to add the charges of the electrons and protons together. For (a), the net charge would be -5 x 10^14 electrons. For (b), the net charge would be 3 x 10^13 electrons (7 x 10^13 protons + 4 x 10^13 electrons).

2) (a) To find the electrostatic force, we can use Coulomb's Law: F = k(q1q2)/r^2. Plugging in the values, we get F = 8.99 x 10^9 * (12 x 10^-9 * (-18 x 10^-9)) / (0.3)^2 = -2.39 x 10^-3 N. The negative sign means the force is attractive.

(b) When the spheres are connected by a conducting wire, they will reach equilibrium and have the same charge. This means the force between them will be 0.

3) We can use the equation F = qE to find the electric field. Plugging in the values, we get E = F/q = (3.8 x 10^-3 N) / (5 x 10^-9 C) = 7.6 x 10^5 N/C.

4) The electric field created by the proton at the position of the electron in a hydrogen atom would be the same as the field in problem 3, since the proton has the same charge as the electron in magnitude but opposite in sign. The direction of the field would be towards the proton.

5) We can use the equation F = qE to find the electric field. First, we need to find the charge of each electron by using the kinetic energy formula KE = (1/2)mv^2. Plugging in the values, we get q = (2 * 1.6 x 10^-17 J) / (4.8 x 10^5 m/s)^2 = 6.67 x 10^-19 C. Now, we can solve for the electric field: E = F/q = (1.6 x 10^-17 J / 0.1 m) / (6.67