Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be tre
1 answer:
Answer:
2. Smaller value of q = 1/4Q
3. Larger value of q = 3/4Q
Explanation:
1. Please check the attachment for the solution of part 1. Due to time factor I couldn't type out the solution properly.
2. The smaller value is
F = 75% = 75/100
= 3/4 * Q²/16piEod² ---3
When we equate 1 and 3
We get 3Q²/16 = Qq - q² from here we cross multiply to get
3Q² = 16(Qq-q²)
3Q² = 16Qq-16q²
16q² - 16Qq + 3Q² = 0
When solve this out using the quadratic equation formula, we have:
Q/2 +- 2Q/8
We can get the smaller value of q as
Q/2-2Q/8
Solve using LCM we get
2Q/8 = 1/4Q = Q/4
C. The larger value of q
Q/2+2Q/8
= 6Q/8
= 3Q/4
Please use the attachment it will guide you to understand this better.
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where f is the focal length of the mirror,
the distance of the object from the mirror, and 
the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for by using the numbers given in the text of the problem:
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
where f is the focal length of the mirror,
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.