Ask question

Ask question

All categories

2 years ago

11

A small oil droplet is sprayed between the plates of a parallel plate capacitor. The droplet has a net charge equal to that of 3

electrons and feels a total force equal to 9.6 x 10-15 N. What is the electric fieldbetween the plates of the parallel plate capacitor

1 answer:

masha68 [24]2 years ago

6 0

Answer:

…..

Explanation:

give brainiest

You might be interested in

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$

this expression holds for the points located at

-r₁ <x₁ <r₁

3) Point between the two spheres

E_ {total} = $k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}$

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

E_ {total} = $+ k \frac{Q}{(d-x_1)^2}$ + k Q / (d-x1) 2

point range

-r₂ <x₂ <r₂

5) Right side point

E_ {total} = $k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}$

E_ {total} = $- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )$ - k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0

2 years ago

The buoyant force on a submerged boulder acts upward because _______.

vovikov84 [41]

Greater water pressure acts on the bottom than on the top

4 0

3 years ago

Differences between freezing point and melting point (Atleast 5 differences)

labwork [276]

Answer:

What is freezing point?

A liquid's freezing point is determined at which it turns into a solid. Corresponding to the melting point, the freezing point often rises with increasing pressure. In the case of combinations and for some organic substances, such as lipids, the freezing point is lower than the melting point. The first solid which develops when a combination freezes often differs in composition from the liquid, and the development of the solid alters the composition of the remaining liquid, typically lowering the freezing point gradually. Utilizing successive melting and freezing to gradually separate the components, this approach is used to purify mixtures.

What is melting point?

The temperature at which a purified substance's solid and liquid phases may coexist in equilibrium is referred to as the melting point. A solid's temperature goes up when heat is added to it until the melting point is achieved. The solid will then turn into a liquid with further heating without changing temperature. Additional heat will raise the temperature of the liquid once all of the solid has melted. It is possible to recognize pure compounds and elements by their distinctive melting temperature, which is a characteristic number.

The difference between freezing point and melting point:

While a substance's melting point develops when it transforms from a solid to a liquid, a substance's freezing point happens when a liquid transforms into a solid when the heat from the substance is removed.
When the temperature rises, the melting point can be seen, and when the temperature falls, the freezing point can be seen.
When a solid reaches its melting point, its volume increases; meanwhile, when a liquid reaches its freezing point, its volume decreases.
While a substance's freezing point is not thought of as a distinctive attribute, its melting point is.
While external pressure is a significant component in freezing point, atmospheric pressure is a significant element in melting point.
Heat must be supplied from an outside source in order to reach the melting point for such a state shift. When a material is at its freezing point, heat is needed to remove it from the substance in order to alter its condition.

<em>Reference: Berry, R. Stephen. "When the melting and freezing points are not the same." Scientific American 263.2 (1990): 68-75.</em>

7 0

1 year ago

A seamount is an isolated land mass rising from the ocean floor.<br> a. True<br> b. False

iris [78.8K]

The answer is A, True.

7 0

3 years ago

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

3 0

3 years ago