Which of the following means that an image is located behind a lens? A. +di B. -di C. -do D. +do
1 answer:
For a lens, the following sign convention is generally used:
- f (the focal length) is positive for a converging lens and negative for a diverging lens
-
(the distance of the object from the lens) is positive if the object is in front of the lens and negative if it is behind the lens
-
(distance of the image from the lens) is positive if the image is behind the lens (real image) and negative if the image is in front of the lens (virtual image)
Therefore, the correct option is
<span>A. +di
</span>which mens that the image is real and located behind the lens.
- f (the focal length) is positive for a converging lens and negative for a diverging lens
-
-
Therefore, the correct option is
<span>A. +di
</span>which mens that the image is real and located behind the lens.
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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
