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3 years ago

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I NEED HELP PLEASE, THANKS! Light passes from air into water at an angle of 40.0° to the normal. What is the angle of refraction

? Please show all work.

1 answer:

Vlad [161]3 years ago

8 0

Answer:

Angle of Refraction = 28.9 degrees

Explanation:

<u><em>We'll use Snell's law for this. It's mathematical form is:</em></u>

=> $n_1* sin(\alpha _1)=n_2 * sin(\alpha _2)$

Where $n_{1} = 1, n_{2} = 1.33, \alpha _{2} = 40^o$

=> $n_{1}$ and $n_{2}$ are the refractive indexes of the air and water respectively.

<u><em>Solution:</em></u>

=> 1 * sin (40) = 1.33 * sin( $\alpha _{2}$ )

=> sin $\alpha _{2}$ = $\frac{sin 40^0}{1.33}$

=> sin $\alpha _{2}$ = 0.4821

=> $\alpha _{2}$ = 28.9 degrees

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Complete Question

A 10-cm-long thin glass rod uniformly charged to 8.00 nC and a 10-cm

long thin plastic rod uniformly charged to -8.00 nC are placed side by

side, 4.20 cm apart. What are the electric field strengths E_1 to E_3 at

distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line

connecting the midpoints of the two rods

a.) Specify the electric field strength E1

b.) Specify the electric field strength E2

c.) Specify the electric field strength E3

Answer:

$E_1=7.13*10^5 N/C$

$E_2= 2.95*10^{5} N/C$

$E_3= 3.84*10^5 N/C$

Explanation:

From the question we are told that

The length of the thin glass is $L = 10 cm$

The charge on the glass rod is $q_g = 8.00nC = 8* 10^{-9} C$

The length of the plastic rod is $L_p = 10cm$

The charge on the plastic rod is $q_p =- 8.00nC = -8.0*10^{-9}C$

The distance between the materials is $d = 4.20cm = \frac{4.2}{100} =0.042m$

The various distances to obtain electric field of are $r_1 = 1.0cm$

$r_2 = 2.0cm$

$r_3 = 3.0cm$

The objective of the solution is to obtain the electric field $E_1 , E_2 \ and E_3$ at distance $d_1 , d_2 \ and \ d_3$ from the glass rod along the line connecting its mid point

Generally electric field of a charge rod at a distance of r the line dividing the rod into half is mathematically represented as

$E = k \frac{2Q}{r\sqrt{L^2 + 4r^2} }$

For the $r_2 = 1.0cm = \frac{1}{100} = 0.01m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r \sqrt{L^2 + 4r^2_1} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_1) \sqrt{L^2 + 4r^2_1} }$

The net electric field is,

$E_{net} =E_1= E_l + E_r$

$= k \frac{2Q}{r_1\sqrt{L^2 + 4 r^2_1 } } + k \frac{2Q}{(0.04-r_1) \sqrt{L^2 + 4 (0.044 -r_1)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.01)\sqrt{(0.01^2 + 4(0.01)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.01)\sqrt{(0.01)^2 + (4) (0.042 - 0.01)^2} }$

$= 6.44*10^5 + 6.9*10^4$

$E_1=7.13*10^5 N/C$

For the $r_2 = 2.0cm = \frac{2}{100} = 0.02m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r_2 \sqrt{L^2 + 4r^2_2} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_2) \sqrt{L^2 + 4r^2_2} }$

The net electric field is,

$E_{net} =E_2= E_l + E_r$

$= k \frac{2Q}{r_2\sqrt{L^2 + 4 r^2_2 } } + k \frac{2Q}{(0.04-r_2) \sqrt{L^2 + 4 (0.044 -r_2)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.02)\sqrt{(0.02^2 + 4(0.02)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.02)\sqrt{(0.02)^2 + (4) (0.042 - 0.02)^2} }$

$= 1.6*10^{5}+ 1.3*10^{5}$

$E_2= 2.95*10^{5} N/C$

For the $r_3 = 3.0cm = \frac{3}{100} = 0.03m$

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

$E_l = k \frac{2Q }{r_3 \sqrt{L^2 + 4r^2_3} }$

The electric filed by the positively charge glass rod on the right side of the dividing line is mathematically represented as

$E_r = k \frac{2Q }{(0.044 - r_3) \sqrt{L^2 + 4r^2_3} }$

The net electric field is,

$E_{net} =E_3= E_l + E_r$

$= k \frac{2Q}{r_3\sqrt{L^2 + 4 r^2_3 } } + k \frac{2Q}{(0.04-r_3) \sqrt{L^2 + 4 (0.044 -r_3)^2} }$

Where k is know as the coulomb's constant with a constant value of

$k = 9*10^9 \ kgm^3 s^{-4} A^{-2}$

$=(9*10^9) \frac{(2) (8*10^{-9})}{(0.03)\sqrt{(0.03^2 + 4(0.03)^2)} } + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.03)\sqrt{(0.03)^2 + (4) (0.042 - 0.03)^2} }$

$= 7.2 *10^{4} + 3.1*10^5$

$E_3= 3.84*10^5 N/C$

8 0

3 years ago