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2 years ago

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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

ated as particles and are fixed with a certain separation. (a) For what value of q/Q will the electrostatic force between the two spheres be maximized? What are the (b) smaller and (c) larger values of q/Q that give a force magnitude that is 75% of that maximum?

1 answer:

forsale [732]2 years ago

7 0

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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Complete Question

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Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

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$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

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