Electric Field & Work Conceptual Questions

[exitwound]

Homework Statement

30. (a) Given the electric field of the earth: -150 N/C k, determine the potential difference from the floor to the ceiling, a distance of 3.0 m. (b) As a positive charge moves from the floor to the ceiling does it gain or lose potential energy? (c) As a positive charge moves from the floor to the ceiling does the electric field do positive or negative work? (d) How about a negative charge?

Homework Equations

$$V_f-V_i=\int \vec{E}\cdot d\vec{s}$$

The Attempt at a Solution

a.) $$\Delta V=-Ed$$

$$\Delta V=150(3)=450V$$

b.)Electric Potential is electrostatic potential energy per unit charge, so if the potential difference becomes positive, the potential energy of the positive charge should become more positive as well. Correct? Anyone enlighten me with a better answer?

c.) (I've always had trouble with energy problems. I can't wrap my head around how objects can be related to PE and KE.) In this case, The work that has to be done to move a charge from floor to ceiling would be the amount of change of potential energy, correct? I don't understand why the answer is negative though. Anyone help on this one?

d.) Answers should be opposite of the positive charge, but I don't understand why.

 

[jeff1evesque]

a.) $$\Delta V=-Ed$$

$$\Delta V=150(3)=450V$$

The equation should be $$\Delta V=Ed$$, but it looks like you did your calculations correctly.

b.)Electric Potential is electrostatic potential energy per unit charge, so if the potential difference becomes positive, the potential energy of the positive charge should become more positive as well. Correct? Anyone enlighten me with a better answer?

Electric potential is defined by $$V \equiv \frac{U_{q}}{q},$$ where $$U_{q + sources}$$ is the potential energy. Potential difference is defined as $$\Delta V = V_{f} - V_{i} \equiv (\frac{U_{q}}{q})_{f} - (\frac{U_{q}}{q})_{i}.$$ So it should mathematically mean that as the potential difference increases, the term $$(\frac{U_{q}}{q})_{f}$$ should be much larger than $$(\frac{U_{q}}{q})_{i}$$, which corresponds to the potential energy $$(U_{q})_{f}$$ to be larger (or $$q$$ really small ).

c.) (I've always had trouble with energy problems. I can't wrap my head around how objects can be related to PE and KE.) In this case, The work that has to be done to move a charge from floor to ceiling would be the amount of change of potential energy, correct? I don't understand why the answer is negative though. Anyone help on this one?

Work to move something say in the gravitational field will yield to a certain amount of potential energy stored in the object. But it may have required more energy than it stored. If you think of throwing a shot-put, you throw further when it's a 45degree angle (compared to any other angles). If you want to throw as far as you did at the 45 degree angle at some other angle- it would require more energy. So I think you could say it's path dependent.

The concept of the conservation laws may help. In particular the conservation of energy states,
$$K_f + qV_f = K_i + qV_i \Leftrightarrow K_F = K_i - q \Delta V.$$ If you move an element (say a proton), it will have some initial velocity, and with other known values (positive charge e) you can solve for $$\Delta V$$. But $$\Delta V$$ has the relationship mentioned above, and depending on what you are looking for, you can simply use the equations to see what's going on. Also an electron has a negative charge $$-e$$, while protons have positive charge $$e$$. Depending on which particle you have, the equations I mentioned would be influenced respectively.

I think I kind of strayed away from your question. The main equation to remember is that,
$$W = F\Delta s = q \vec{E}d$$. So if you have a proton, q will be positive. Now depending on which way the electric field is oriented, this will determine whether the work done is positive or negative. If your charge goes against the electric field, then it will be positive work (since you have to fight the field to get where you're trying to go).

d.) Answers should be opposite of the positive charge, but I don't understand why.

The answer is the opposite because one charge is going along the electric field while another charge is fighting against the electric field. Think of an electron trying traveling towards the negative side of a capacitor plate (work is required), and the converse- an electron going towards a positive plate (the field is acting, and no/little work is required).

I hope that helps, I'm a student as well, and not exactly the sharpest at these things,JL

 

[jeff1evesque]

I made a typo (which I fixed- about work), and I hope I don't have too many. But from my knowledge, I think that's what I know. If anyone else could contribute to help your understanding, that may be beneficial also.

Good Luck,JL