A vector has an x-component

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ .

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Explain how an alpine glacier can change the topography of a mountainous area

igor_vitrenko [27]

An alpine glacier can change the topography of a mountainous area through Glacial Erosion and Glacial Deposition. Glaciers are agents of erosion, it can pick up and carry large rocks and sediments. In the process, a deep cavity or hole can form when the glacier plucks a big rock from where it passed. Glaciers have shaped many Mountain Ranges and have created distinct landforms by its erosion process. In Glacial Deposition, as glaciers melt, it deposits all that it carried and a landform is developed.

7 0

2 years ago

If

nadya68 [22]

Answer:

hhjjkkkksksksjskkskakakkskskksksksoao

Explanation:

hiiii look forward but I don't know how to do it

8 0

3 years ago

Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it

MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r = r-h / r

2h = r-h

3h = r

h = r / 3

5 0

3 years ago

(4.3 x 10-2 )4.950 x 105) O 21.285 x 103 O 2.13 x 104 O 21 x 103 O 2.1 x 104

Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0

2 years ago