All categories

3 years ago

Which law describes the interactions between charged particles when they are not in contact?

Physics

2 answers:

natulia [17]3 years ago

7 0

The answer would be the coulombs law.

mojhsa [17]3 years ago

7 0

Answer:

B) Coulomb's law

Explanation:

As per coulombs law if two point charges are placed at some distance from each other then the force of interaction between them is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 , q_2$ = two charges between which force of interaction is required

d = distance between two charges

$k = \frac{1}{4\pi \epsilon_0}$

Faraday law is used to find the induced EMF due to changing magnetic flux

Ohm's law is used to relate potential difference between the ends of a conductor with its resistance and its current

You might be interested in

Which best defines the skeletal system?

slamgirl [31]

I’d say B is the best choice

Hope this helps

-AaronWiseIsBae

6 0

3 years ago

Have I answered this question correctly?

REY [17]

I believe so sorry if you didn’t lakalalallalalalalallalala

5 0

3 years ago

How many coffee beans selling for $1.20 per pound should be mixed with 3 pounds of coffee beans selling for $2.40 per pound to o

Ulleksa [173]

This is not really a physics question but I can still help. It’s easier if you make a table. I hope this helped.

8 0

3 years ago

A blue and a green billiard ball, each with a mass of 0.15 kg, collide directly. Before the collision, the blue ball had a speed

cestrela7 [59]

Answer:

Explanation:

Can you specify what is the question asking for

4 0

2 years ago

A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with

Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

$q=1.558*10^{-9}c$

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength $E_l=784.75N/m$

Right field strength $E_r=776.38 N/m$

Front field strength $E_f=725.5 N/m$

Back field strength $E_b=749.54 N/m$

Top field strength $E_t=944.95 N/m$

Bottom field strength $E_{bo}=1082.58 N/m$

Generally, the equation for Charge flux is mathematically given by

$\phi=EAcos\theta$

Where

Theta for Right,Left,Front and Back are at an angle 90

$cos 90=0$

Therefore

$\phi =0$ with respect to Right,Left,Front and Back

Generally, the equation for Charge Flux is mathematically also given by

$\phi=\frac{q}{e_o}$

Where

$Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2$

Therefore

$Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t$

$Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)$

$Q_{net}=176N/C m^2$

Giving

$q=\phi*e_0$

$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

4 0

3 years ago