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3 years ago

A positively charged light metal ball is suspended between two oppositely charged metal plates on

Physics

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

The positively charged ball moves between both charged plates till the plates and the ball all become neutral.

Check Explanation for more.

Explanation:

Let the ball be in square brackets, and the plates in normal brackets.

(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.

The ball hanging on the insulated thread becomes neutral too at this point.

Hope this Helps!!!

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3 years ago

A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc

balandron [24]

Answer:

Approximately $28^{\circ}$ .

Explanation:

The refractive index of the air $n_{\text{air}}$ is approximately $1.00$ .

Let $n_\text{glass}$ denote the refractive index of the glass block, and let $\theta _{\text{glass}}$ denote the angle of refraction in the glass. Let $\theta_\text{air}$ denote the angle at which the light enters the glass block from the air.

By Snell's Law:

$n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})$ .

Rearrange the Snell's Law equation to obtain:

$\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}$ .

Hence:

$\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}$ .

In other words, the angle of refraction in the glass would be approximately $28^{\circ}$ .

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