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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

4 0

3 years ago

Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman

Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2$

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0

3 years ago

Viewed from a distance, how would a flashing red light appear as it fell into a black hole?.

Pachacha [2.7K]

Explanation:

Light rays that pass close to the black hole get caught and cannot escape. so what ends up happening to the light is that it starts bending due to the strong gravitational force of the black hole.

5 0

2 years ago

Se deja caer un objeto desde la cornisa de una casa la cual tarda en llegar al suelo 645/500 segundos. ¿Cúal es la altura de la

salantis [7]

8.15m. La altura de la casa de la cual se deja caer un objeto que tarda en caer $\frac{645}{500}s$ es de 8.15m.

La clave para resolver este problema es mediante caída libre, la velocidad inicial del objeto es 0 por lo que podemos calcular la altura h de la casa mediante la relación $h= \frac{gt^{2}}{2}$ .

Conocemos la gravedad $g=9.8\frac{m}{s^{2}}$ y el tiempo en que tarda en caer el objeto $t=\frac{645}{500}s$ , sustituyendo los valores tenemos que:

$h=\frac{(9.8\frac{m}{s^{2}})(\frac{645}{500}s)^{2}}{2}=8.15m$

4 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

Genrish500 [490]

Explanation:

Given:

Final speed of mass A = Va

Final speed of mass B = Vb

Mass of A = Ma

Mass of B = Mb

Ma = 2 × Mb

By conservation of linear momentum,

0 = Ma × Va + Mb × Vb

0 = 2 × Mb × Va + Mb × Vb

Vb = - 2 × Va

Energy of the spring, U = 1/2 × k × x^2

1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²

35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²

Ma × Va² + Mb × Vb² = 70

2 × Mb (-Vb/2)² + Mb × Vb² = 70

1/2 × Mb × Vb² + Mb × Vb² = 70

3/2 × Mb × Vb² = 70

Mb × Vb² = 140/3

= 46.7 J

Ma = 2 × Mb and Vb = - 2 × Va

Ma/2 × (4 × Va²) = 140/3

Ma × Va² = 70/3

Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J

Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J

8 0

3 years ago