Can an electric field exist in a vacuum?

The distance and direction between starting and ending positions

rodikova [14]

Answer:

The length traveled by an object moving in any direction or even changing direction is called distance. The location of an object in a frame of reference is called position. For straight line motion, positions can be shown using a number line. The separation between original and final position is called displacement

Explanation:

good luck

8 0

3 years ago

The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. Use the average of the l

lana [24]

Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

From the left we have:

r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.

And this should be the correct answer. Watch your chart and replace if it's neccesary.

3 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

Which of the following is an example of kinetic energy being converted to potential energy?

horsena [70]

Of the following...?? Is there more to this question? :)

4 0

2 years ago

Read 2 more answers

What is solar wind i ask that you help me

Novay_Z [31]

My answer -

the corona, the sun's outer layer, reaches temperatures of up to 2 million degrees Fahrenheit (1.1 million Celsius). At this level, the sun's gravity can't hold on to the rapidly moving particles, and it streams away from the star.

The sun's activity shifts over the course of its 11-year cycle, with sun spot numbers, radiation levels, and ejected material changing over time. These alterations affect the properties of the solar wind, including its magnetic field properties, velocity, temperature and density. The wind also differs based on where on the sun it comes from and how quickly that portion is rotating.

The velocity of the solar wind is higher over coronal holes, reaching speeds of up to 500 miles (800 kilometers) per second. The temperature and density over coronal holes are low, and the magnetic field is weak, so the field lines are open to space. These holes occur at the poles and low latitudes, and reach their largest when activity on the sun is at its minimum. Temperatures in the fast wind can reach up to 1 million degrees F (800,000 C).

At the coronal streamer belt around the equator, the solar wind travels more slowly, at around 200 miles (300 km) per second. Temperatures in the slow wind reach up to 2.9 million F (1.6 million C).

p.s

Glad to help you and if you need anything else on brainly let me know so I can elp you again have an AWESOME!!! :^)

4 0

2 years ago