Question

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

Answer 1

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

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

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

        $= 7.2 *10^{4} + 3.1*10^5$

      $E_3= 3.84*10^5 N/C$