Question

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q2 = +1.00 µC, and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) r = 4.00 m, (b) r =0.700 m, and (c) r = 0.200 m? With V = 0 at infinity, what is V at (d) r = 4.00 m, (e) r = 1.00 m, (f) r = 0.700 m, (g) r = 0.500 m, (h) r = 0.200 m, and (i) r = 0? (j) Plot the E(r) and V(r) dependencies.

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =$1.685*10^3 N/C$

b E =$36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e   $V = 26.79*10^3V$

f   V = $34.67 *10^3 V$

g   $V= 44.95*10^3 V$

h    $V= 44.95*10^3 V$

i    $V= 44.95*10^3 V$

Explanation:

From the question we are given that

       The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

       The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

      The first radius $R_1 = 0.500m$

      The second radius $R_2 = 1.00m$

 $Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           $E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

                $= 1.685*10^3 N/C$

Considering b

           $r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

            $E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

             This because it the electric filed of the charge which is below it in distance that it would feel

            $E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

               = $36.69*10^3 N/C$

   Considering c

                      r  = 0.200 m

=>   $r$

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

                  $V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m $R_2 = r > R_1$

                $V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

          Considering f

              $r = 0.700 m \ , R_2 > r > R_1$

                      $V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

          Considering g

             $r =0.500\m , R_1 >r =R_1$

   $V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

          Considering h

                $r =0.200\m , R_1 >R_1>r$

  $V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

           Considering i    

   $r =0\ m \ , R_1 >R_1>r$

  $V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$