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Ganezh [65]
3 years ago
12

When applying Gauss' law for a capacitor containing a dielectric, which of the following statements is false? (Latter part of se

ction 6.)
A) The electric field vector is multiplied by κ, the dielectric constant.
B) The charge enclosed by the Gaussian surface includes both the free charge and the induced surface charges.
C) Gauss' law for a dielectric includes the ability to account for varying dielectric constant over the Gaussian surface.
D) Gauss' law for a dielectric is the most general form of Gauss' law.
E) Gauss' law for a dielectric can be applied to many types of capacitors.
Physics
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:a

Explanation:

Introducing a dielectric inside a capacitor reduces the Electric field . Charge develops on dielectric such that it is opposite to the corresponding face . For example Positive charge of capacitor induces negative charge on dielectric which will create an Electric Field in opposite direction.

So Electric Field decreased by a factor of \frac{1}{K} where

K=dielectric\ magnitude    

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A laboratory electromagnet produces a magnetic field of magnitude 1.50T . A proton moves through this field with a speed of 6.00
hram777 [196]

The magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

The magnitude of the force on a charged particle moving in a magnetic field is given by the formula,

F= qvB

Here q is the charge on proton = 1.6 x 10^-19 C.

v is the velocity with which the particle is moving = 6.00 x 10^6 m/s

And B is the value of the magnetic field = 1.5 T

Putting the given values in the above equation,

F = 1.6 x 10^-19 x 6 x 10^6 x 1.5 = 1.44 x 10^-12 N.

Hence, the magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

To know more about "magnetic force", refer to the link given below:

brainly.com/question/13791875?referrer=searchResults

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3 0
1 year ago
1. Why should you use scientific notation when writing big numbers?
Fynjy0 [20]

Answer:

It helps the answer look clean. It also makes the work easier to work with.

Explanation:

Instead of writing a lot of zeros, all you have to do is add exponents to the number to show how much the decimal moved.

6 0
3 years ago
The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e
FrozenT [24]

Answer:

1.1486813808\times 10^{-7}\ T

34.46 V/m

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

c = Speed of light = 3\times 10^8\ m/s

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T

The magnetic field intensity is 1.1486813808\times 10^{-7}\ T

The maximum electric field intensity is given by

E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m

The  electric field intensity is 34.46 V/m

8 0
2 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
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