Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor,
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
e is the charge on electron
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
Explanation:
(i)
O is the object and I is the image.
The image formed is enlarged and it is erect. So the magnification will be positive (+) and greater than 1.
Refer above image. 1
(ii)
O is the object and I is the image.
The image formed is diminished and erect. So the magnification will be positive (+) and less than1.
Refer above image. 2
(iii)
The image will be formed as the 2F on the other side of the lens and it will be of same of the object.
Answer: a ) 6/19 V/L
b ) 3/19
Explanation:
