Physics, 20.04.2020 23:30 angie3892

A 10-cm-long thin glass rod uniformly charged to 8.00 nCnC and a 10-cm-long thin plastic rod uniformly charged to - 8.00 nCnC are placed side by side, 4.20 cmcm apart. What are the electric field strengths E1E1E_1 to E3E3E_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

Answers: 3

Show answers

Answers

Answer from: alayciaruffin076

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

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

Answer from: Quest

Iam looking for this answer too!

Answer from: Quest

Mfjh f fb gcbb etyb yryh htbgh g yrjjj t n j tj

Another question on Physics

Physics, 22.06.2019 03:30

The solar panels used by mark function because of the photoelectric effect. light shines on the cells causing electrons to be ejected from the metal, which produces an electric current. at night on mars, no light will fall on the solar cells and no electric current will be generated. according to your notes, what type of light is typically needed to cause the photoelectric effect? a)visible b)ultraviolet c)infrared

Answers: 1

Answer

Physics, 22.06.2019 14:30

What is the relationship between the direction of motion of the balloon and the wind currents

Answers: 1

Answer

Physics, 22.06.2019 20:10

Consider two less-than-desirable options. in the first you are driving 30 mph and crash head-on into an identical car also going 30 mph. in the second option you are driving 30 mph and crash head-on into a stationary brick wall. in neither case does your car bounce off the thing it hits, and the collision time is the same in both cases. which of these two situations would result in the greatest impact force?

Answers: 1

Answer

Physics, 23.06.2019 12:00

What was the approximate temperature of the universe at an age of 1015 s?

Answers: 3

Answer

You know the right answer?

A 10-cm-long thin glass rod uniformly charged to 8.00 nCnC and a 10-cm-long thin plastic rod uniform...

Questions