All categories

3 years ago

The diagram does not represent a real electric field because the field lines, can someone help explain this for me

Physics

1 answer:

Papessa [141]3 years ago

4 0

electric field lines are graphical presentation of electric field intensity

It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

here we will always follow these basic properties of field lines

now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

As we have two directions of tangents at that point

So here the incorrect presentation is the intersection of two field lines which is not possible

You might be interested in

(Fill the blank) the ________ g____ makes saliva that make your food wet and easy to swallow

blagie [28]

Answer:

What make saliva that make your food wet and easy to swallow?

enzyme amylase

Explanation:

The digestive functions of saliva include moistening food, and helping to create a food bolus, so it can be swallowed easily. Saliva contains the enzyme amylase that breaks some starches down into maltose and dextrin. Thus, digestion of food occurs within the mouth, even before food reaches the stomach.

Hope it help 4u

Yan na baby

6 0

2 years ago

पत्र- अपने क्षेत्र की सड़कों की बुरी दशा की जानकारी देते हुए और उन्हें ठीक कराने की प्रार्थना करते हुए नगर निगम अधिकारी को पत्र

Ksivusya [100]

Answer:

पत्र- अपने क्षेत्र की सड़कों की बुरी दशा की जानकारी देते हुए और उन्हें ठीक कराने की प्रार्थना करते हुए नगर निगम अधिकारी को पत्र लिखिए |

5 0

2 years ago

What are the different types of topology?

dolphi86 [110]

Answer:

common types of topologies, and we're going to break each of them down in the guide below.

Bus topology. As the simplest design, a bus topology requires nodes to be in a linear order. ...

Ring topology. Another simple design is the ring topology. ...

Star topology. ...

Mesh topology. ...

Tree topology.

Explanation:

3 0

1 year ago

Read 2 more answers

A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu

Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

= 25° C

The specific heat capacity of aluminium, c = 900 J/kg°C

The formula for thermal energy,

Q = mcΔT

= 1.0 x 900 x 25

= 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago