As a result of fringing effect :

What is the electric field strenght in units of N/C if the flux through a 2.0m by 1.0m rectangular surface is 396Nm2/C, if the e

Elis [28]

Answer:

The appropriate answer is "396 N/C".

Explanation:

The give values are:

Rectangular surface,

$\Phi_e=396 \ N.m^2/C$

$a = 2 \ m$

$b = 1 \ m$

Angle,

$\theta =\frac{\pi}{3} \ radian$

Now,

The area of rectangle (A) will be:

= $a\times b$

= $2\times 1$

= $2 \ m^2$

hence,

The electric field strength will be:

⇒ $\Phi_e=E.A Cos \theta$

or,

⇒ $E=\frac{\Phi_e}{ACos \theta}$

On putting the given values, we get

⇒ $=\frac{396}{2\times Cos(\frac{\pi}{3} )}$

⇒ $=\frac{396}{2\times 0.5}$

⇒ $=\frac{396}{1}$

⇒ $=396 \ N/C$

6 0

3 years ago

This is due by 11:59 PM tonight.

svetoff [14.1K]

Answer:

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

6 0

4 years ago

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 490 nm occurs at an angl

Black_prince [1.1K]

Answer:

(a) d = 1960nm

(b) The slit should be decreased.

(c) Δd = 360nm.

Explanation:

The double-slit interference is given by the following equation:

$d sin(\theta) = m \lambda$ (1)

where d: is the distance between slits, Θ: is the angle between the path of the light and the screen, m: is the order of the interference and λ: is the wavelength of the light.

(a) To determine the least wavelength in the visible range in the third-order we need first to find the distance between slits, using equation (1) for a fourth-order:

$d = \frac{m \lambda}{sin(\theta)} = \frac{4 \cdot 490nm}{sin(90)} = 1960nm$

Now, we can find the least wavelength in the visible range in the third-order:

$\lambda = \frac{d sin(\theta)}{m} = \frac{1960nm sin(90)}{3} = 653nm$

So, the least wavelength in the visible range (400nm - 700nm) in the third-order is 653nm.

(b) To eliminate all of the visible light in the fourth-order maximum means that the wavelength must be smaller than 400nm, and hence the slit separation should be decreased since they are proportional to each other (see equation (1)).

(c) The distance between slits needed to eliminate all of the visible light in the fourth-order maximum, with λ = 400 nm as limit value, is:

$d = \frac{m \lambda}{sin(\theta)} = \frac{4 \cdot 400nm}{sin(90)} = 1600nm$

Therefore the least change in separation needed is equal to the initial distance calculated for 490nm and the final distance calculated for 400nm:

$\Delta d = d_{i} - d_{f} = 1960nm - 1600nm = 360nm$