I need someone who is great at physics to please help me with these 6 questions.
ID: 1769923 • Letter: I
Question
I need someone who is great at physics to please help me with these 6 questions...I dont understand any of them!
1. How much work would be required to move an electron from one equipotential to the adjacent one at a lower potential? Indicate which two lines you are using.(this is where numbering lines is useful)
2. Calculate the average electric field along a section of a field line between the two equipotential lines.
3. What average force would be needed to move the electron from one of these equipotentials to the other?
4. If the DC voltage source was halved to 4V or doubled to 16V would the shape of the equipotential lines be different or the same? Why?
5. Calculate the magnitude of the electric field in three different locations for the point plate configuration. does the magnitude of the field vary as you would expect it to in each of the configurations?
6. On your sketch for point plate configuration locate the region of your map where the electric field is the weakest and mark its location. explain how you determine the location. explain how you determined the direction of the electric field from your map.
t/, 2N 91012 14 16 18 1 2022 2426Explanation / Answer
1. suppose i use 5V and 3V line in figure 3.
Let v(a) = 5V and v(b) = 3V. Then work done in moving the electron from 5V line to 3V line is w = e(v(a)-v(b))=1.6x10^(-19)x(5-3) = 3.2 x 10^-19 Joule.
2. between the same two equipotential lines as in part 1, the distance between these two lines along the red line in the figure is d = 6 cm =0.06m (i assume that distance is in cm, although no unit for distance is mentioned in the question. Then electric field beween these two lines along the red line E = (v(a)-(v(d)) / d = (5-3) / .06 = 100/3 =33.3 V/m from 5V line to 3V line as electric field is from higher potential to lower potential.
3. Let F be the force required to move an electron from higher potential to lower potential line between the same two lines as we have used in part 1. and part 2.
Work required to be done W = eV = Fd
so F X .06 = (1.6 x 10^-19) X 2
F = (3.2 x 10^-19)/.06 = 5.3 x 10^-18 N.
4. If the DC source was halvedto 4V then the spacing between the equipotential surfaces will be doubled as electric field is halved because elctric field in magnitude is equal to the gradient of potential. If DC source is doubled to 16V then spacing between the equipotential surfaces will be halved due to same reasons.
for part 5 you consider any two equipotential surfaces and measure distance between them at any three different locations and find potential gradient by dividing potential difference by distance and please note that electric field at any point is perpendicular to equipotential at every point.
For part 6. please note that electric field is weakest where the spacing between the equipotential surfaces is maximum and is from higher potential to lower potential.