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3 years ago

When you double the distance between a pair of charged particles what happens to the force between them?

Physics

1 answer:

Volgvan3 years ago

8 0

Answer:

Force between the two charges becomes one fourth of the initial force.

Explanation:

The electrostatic force acting between any two charges is given as,

$F = k\frac{q_{1}q_{2}}{r^{2}}$

Here,

F = force

k = Coulomb's constant

$q_{1}$ = magnitude of charge of the first particle

$q_{2}$ = magnitude of charge of the second particle

$r$ = separation between the two charges

From the above relation,

$F \propto \frac{1}{r^{2}}$

Thus,

$\frac{F_{1}}{F_{2}} = \left ( \frac{r_{2}}{r_{1}} \right )^{2}$

$\frac{F_{1}}{F_{2}} = \left ( \frac{2r}{r} \right )^{2}$

$\Rightarrow \ F_{2} = \frac{1}{4}F_{4}.$

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3 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

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a) Constant

b) Constant

Explanation:

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

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