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Fudgin [204]
3 years ago
6

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

the Coulomb force between the two is to be maximized?
Physics
1 answer:
Lisa [10]3 years ago
6 0

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

F = \frac{Kq_{1}q_{2}}{r^{2}}

So, the force between the charges q and Q - q is given by

F = \frac{K\left ( Q-q \right )q}}{r^{2}}

For maxima and minima, differentiate the force with respect to q.

\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now \frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}

the double derivate is negative, so the force is maxima when q = Q / 2 .

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Answer: The refractive index or n shows the relationship between c and the speed v when light travels through a material. The equation is n=c/v. In water, the refractive index is 1.3 and in glass, it is 1.5. Therefore, light travels faster through water, than it travels through glass.

Explanation:

4 0
3 years ago
The smallest unit of an element that has all of the properties of the element is a/an
AysviL [449]

Answer:B

Explanation: an atom is the smallest particle of an element that can take part in a chemical reaction.

7 0
3 years ago
A car with a constant velocity of 22 m/s is driven for 6.8 s. How far did it travel? (V =A d/AT)
SVEN [57.7K]

Answer: 150 m

Explanation:

velocity = 22 m/s

t = 6.8 s

velocity = \frac{distance}{time}

distance = velocity × time

= 22 × 6.8

= 149.6 m

≈ 150 m

option C

6 0
3 years ago
Three diffrent examples of accelerated motion
LekaFEV [45]

Answer:

The three different examples of the accelerated motion are Falling/dropping of ball, Standing in circular rotating space, moving around the circle.

Explanation:

Acceleration is the change in velocity, which is related to the speed and direction in which the object is travelling. Hence, speeding up, slowing down and turning are few types . A simple example would be dropping a ball: as it falls its speed increases, which is a type of acceleration. A more complicated example would be standing in a circular, rotating space station. A point on the station moves in a circle, meaning that as it travels it must be turning (to remain in circular motion) making this another example of acceleration

3 0
3 years ago
The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in
bezimeni [28]

1) Potential difference: 1 V

2) V_b-V_a = -1 V

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

\Delta U=q\Delta V

where

q is the charge's magnitude

\Delta V is the potential difference between the initial and final position

In this problem, we have:

q=4.80\cdot 10^{-19}Cis the magnitude of the charge

\Delta U = 4.80\cdot 10^{-19}J is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

V_b-V_a = - 1V

8 0
3 years ago
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