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goldfiish [28.3K]
2 years ago
13

What is the field outside the capacitor plates in a parallel capacitor?​

Physics
2 answers:
FrozenT [24]2 years ago
7 0

Answer is zero

Plz mark me brainlist

mr Goodwill [35]2 years ago
7 0

Explanation:

Outside two infinite parallel plates with opposite charge the electric field is zero, and that can be proved with Gauss's law using any possible Gaussian surface imaginable

You might be interested in
Fe₂O3<br> + co<br> →<br> Fe3O4 + CO₂
Goryan [66]

Explanation:

                    Fe₂O₃  + CO  → Fe₃O₄ + CO₂

Balancing the equation above, we can derive simple mathematical equations that are very easy to solve.

             aFe₂O₃  + bCO  → cFe₃O₄ + dCO₂

a,b,c and d are the coefficients needed to balance the equation above;

  Conserving Fe; 2a = 3c

                       O: 3a + b = 4c + 2d

                        C: b = d

 let a = 1;

      c = \frac{2}{3}

      Since b = d

                  3a + d = 4c + 2d

                    3a = 4c + 2d - d

                     3a = 4c + d

           a = 1, c = \frac{2}{3}

                    3 = 4 x \frac{2}{3}  +  d

                   d = \frac{1}{3}

                    b = \frac{1}{3}

multiplying a, b, c and d by 3:

            a = 3    b = 1     c = 2   and d = 1

                  3Fe₂O₃  + CO  → 2Fe₃O₄ + CO₂

Learn more:

Balanced equation brainly.com/question/2612756

#learnwithBrainly

6 0
3 years ago
2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

5 0
3 years ago
Help me with my physics, please
Readme [11.4K]
The right answer would be

-20t+ 80
7 0
3 years ago
A box slides down a frictionless incline, gaining speed. The work done by the normal force n is _______.
jeka57 [31]

The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

  • The body is gaining the speed, which means there is a change in kinetic energy.
  • The change in kinetic energy is equal to the work done.
  • The friction force is the product of coefficient of the friction and normal force.
  • The friction force for the given case is zero. Thus, the normal force must be equal to the zero.

Thus, the work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

Learn more about the normal force here;

brainly.com/question/10941832

7 0
2 years ago
Read 2 more answers
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