course phys 202
Ass 261.Based on what you learned from Introductory Problem Set 2, how does the current in a wire of a given material, for a given voltage, depend on its length and cross-sectional area
- (V / L2) / (V / L1) = L1 / L2 needs the length in order to find the ratio
2. How can the current in a wire be determined from the drift velocity of its charge carriers and the number of charge carriers per unit length?
- (V / L2) / (V / L1) = L1 / L2 one can use the same equation then the ratio of L1/L2 is the drift ratios which is also equal to the ratio potential gradients
3. Will a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why?
- greater area implies proportionally more available charge carriers per unit of length and hence proportionally more current.
4. 4. Will a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why
- I believe the electrical resistance will be greater in the smaller cross-sectional area tube, so the drift potential gradient will be higher too, which means a higher current
If length is the only difference, then greater length implies a lesser electric field and therefore a lesser drift velocity and a lesser current.
5. Force on proton is 3.75 * 10^-14 N toward south. What is magnitude and direction of the field
- the magnitude is equal to the the mag. of force/the charge.
=3.75 * 10^-14 N / (1.6 * 10^-19 C)
= 2.36* 10^5 N / C
7. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge?
- we can use k q / r^2 and r=.08m
- since they are equally giving they have an magnitude of .5(745)=
373 N/C
-then we solve for q, after figuring what variables we have, so
q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2
= 2.6 * 10^-10 C
8. If the charges are represented by Q and -Q, what is the electric field at the midpoint?
- the electric field at the midpoint would be found using 2 k Q / r^2, using .08m as the radius again and the factor is 2 because there is two charges
9. Electric field 20.0 cm above 33.0 * 10^-6 C charge?
- the force is going to upward so we use the equation E = (k q Q / r^2) / Q, after substituting we get
E = 9 * 10^9 N m^2 / C^2 * 33.0 * 10^-6 C / (.200 m)^2
= 7.43 * 10^6 N / C
&#See my notes and let me know if you have questions. &#