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STALIN [3.7K]
1 year ago
9

Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl

e of charge q as the particle travels from A to B is WAB and that done on the particle as it travels from A to C is WAC=−WAB/3.
a) How much electrostatic work is done on the particle as it travels from B to C?
Physics
1 answer:
Trava [24]1 year ago
4 0

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac =  q (  Va -  Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]  

= Wac - Wab

= Wac + 3Wac

= 4Wac

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A measuring microscope is used to examine the interference pattern. It is found that the average distance between the centers of
diamong [38]

Answer:

 2n t = m λ₀ ,    R = 0.240 mm

Explanation:

The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.

The ray that is reflected in the upper part of the film has a phase change of 180º since the ray stops from a medium with a low refractive index to one with a higher regrading index,

-This phase change is the introduction of a λ/2 change

-The ray passing through the film has a change in wavelength due to the refractive index of the medium

          λ₀ = λ / n

Therefore Taking into account this fact the destructive interference expression introduces an integer phase change, then the extra distance 2t is

        2 t = (m’+ ½ + ½) λ₀ / n

        2t = (m’+1) λ₀ / n

         m = m’+ 1

        2n t = m λ₀

        With   m = 0, 1, 2, ...

Where t is the thickness of the film, n the refractive index of the medium, λ the wavelength

The thickness of a hair is the thickness of the film t

           2R = t

             R = t / 2

             R = 0480/2

              R = 0.240 mm

3 0
2 years ago
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1
7nadin3 [17]

Answer:

A). σ = 3.823 x 10^{-5} C^{2}/N-m^{2}

B). \sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C). U=10.322 J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

\sigma =k.\varepsilon _{0}.E

where, E is resultant electric field = 1.2 x 10^{6} V/m

           \varepsilon _{0} is permittivity of free space = 8.85 x 10^{-12} C^{2}/N-m^{2}

           k is dielectric constant = 3.6

∴\sigma =k.\varepsilon _{0}.E

                     = 3.6 x 8.85 x10^{-12} x 1.2 x 10^{6}

                    = 3.823 x 10^{-5} C^{2}/N-m^{2}

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )

\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )

\sigma ^{'}=2.76\times 10^{-5} C/m^{2}

C).

Area of the plate, A = 2.5 cm^{2}

                                 = 2.5 x 10^{-4}m^{2}

diameter of the plate, d = 1.8 mm

                                        = 1800 m

∴ Total energy stored in the capacitor

U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad

U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800

U=10.322 J

4 0
2 years ago
2. Heavier football players tend to play on the front line. Why? <br> What law is it?
gogolik [260]

Answer: They are put in front for defense so so they can block the opponents from getting the ball

Explanation:

4 0
1 year ago
Read 2 more answers
What force would be needed to accelerate a 0.040-kg golf ball at 20.0 m/s?
Naily [24]

Answer:

any amount of force will do it as time is not mentioned here

5 0
2 years ago
A river has a steady speed of vs. A student swims upstream a distance d and back to the starting point. (a) If the student can s
andre [41]

he speed of the student relative to shore is

v_ up = v- vs

v _down = v+ vs

The time required to travel distance d upstream
is

t_up = d/ v_up = d/ v- vs

(2)

The time required to swim the same distance d downstream is

t_down = d/ v_down = d/ v+ vs

6 0
1 year ago
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