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Answer:
Figure a. E_net = 99.518 N/C
Figure b. E_net = 177.151 N / C
Explanation:
Given:
- Attachment for figures missing in the question.
- The dimensions for rectangle are = 7.79 x 3.99 cm
- All four charges have equal magnitude Q = 10.6*10^-12 C
Find:
Find the magnitude of the electric field at the center of the rectangle in Figures a and b.
Solution:
- The Electric field generated by an charged particle Q at a distance r is given by:
E = k*Q / r^2
- Where, k is the coulomb's constant = 8.99 * 10^9
Part a)
- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:
E_1 + E_3 = 0
- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:
E_net = E_2 + E_4
E_2 = E_4
E_net = 2*E = 2*k*Q / r^2
- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:
r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )
r = sqrt ( 1.9151*10^-3 ) = 0.043762 m
- Plug the values in the E_net expression developed above:
E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3
E_net = 99.518 N/C
Part b)
- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).
- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.
Hence,
E_net = 2*E_part(a)*cos(Q)
- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:
Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.
- Now, compute the net electric field E_net:
E_net = 2*(99.518)*cos(27.12)
E_net = 177.151 N / C
The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.
The correct answer is option D.
In the given graph, we can deduce the following;
- the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s
The average speed of the ant is calculated as;
The total distance from the graph is calculated as follows;
- first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm
- first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm
- second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
- second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
- third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm
- fourth downward distance from 12 cm to 9 cm = 3 cm
- final horizontal distance from 13 cm to 15 cm = 2cm
The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm
The average velocity is calculated as the change in displacement per change in time.
The displacement is the shortest distance between the start and end positions.
- This shortest distance is the straight line connecting the start and end position. Call this line P
- From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
- Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm
Notice, you have a right triangle, now calculate the length of line P.
↓end
↓
↓ 6cm
↓
start -------------13 cm------------
Use Pythagoras theorem to solve for P.
The average velocity of the ant is calculated as;
Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.
Learn more here: brainly.com/question/589950