Question

A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 21.0 cm from this surface, the potential is 499 V. Calculate the radius of the sphere. Submit Answer Tries 0/32 Determine the total charge on the sphere. Submit Answer Tries 0/100 What is the electric potential inside the sphere at a radius of 4.0 cm? Submit Answer Tries 0/32 Calculate the magnitude of the electric field at the surface of the sphere. Submit Answer Tries 0/32 If an electron starts from rest at a distance of 21.0 cm from the surface of the sphere, calculate the electron's speed when it reaches the sphere's surface.

Answer 1

Answer:

a

The radius is   $r_1 = 0.315m$            

b

The total  charge is   $Q= 2.912*10^{-8}C$

c

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

d

The magnitude of the electric field is  $E= 2641.3 V/m$

e

The velocity is   $v= 1.76 *10^{14} m/s$

Explanation:

From the question we are told that

     The potential is $V_1 = 832 V$

       The radial distance from the sphere is $d = 21.0cm = \frac{21}{100} = 0.21m$

       The potential at the radial distance is $V_2 = 499V$

The potential at the surface of the sphere is mathematically represented as

                     $V = \frac{kQ}{r}$

                    $Vr = kQ$

Where kQ is  a constant what this means that the the charge Q and the coulomb constant do  not change

  This means that

              $V_1 r_1 = V_2 r_2$

Where $r_1$ is the radius of the sphere

     and $r_2$ is the distance  from that point where the second potential was measured to the center of the sphere which is mathematically represented as

             $r_2 = r_1 + d$

Substituting  this into the equation

                      $v_1 r_1 = V_2 (r_1 +d)$

 Now substituting value

                   $832 * r_1 = 499 * (r_1 + 0.21)$

                   $832r_1 - 499r_1 = 104.79$

                   $333r_1 = 104.79$

                       $r_1 = \frac{104.79}{333}$

                           $r_1 = 0.315m$              

From the equation above

          $V = \frac{kQ}{r_1}$

making Q the subject

        $Q = \frac{V r_1 }{k}$

k has a values of $k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}$

       Substituting into the equation

            $Q =\frac{832 * 0.315}{9*10^9}$

               $Q= 2.912*10^{-8}C$

According to  Gauss law  the electric field from  outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

                  $E = \frac{kQ}{r_1^2}$

Substituting values

                 $E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}$

                     $E= 2641.3 V/m$

 According the the law of energy conservation

  The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

         $EPE = V * e$

Where is e is an electron

And Kinetic energy is mathematically represented as

         $KE = \frac{1}{2} m v^2$

From the statement above

          $V_2 e = V_1 e + \frac{mv^2}{2}$

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

            $V_2 e = \frac{mv^2}{2}$

The charge an electron has a value  $e = 1.602*10^{-19}C$

And the mass of an electron is $m = 9.109 *10^{-31}kg$

     Making v the subject

       $v = \sqrt{\frac{2V_2 e}{m} }$

Substituting value

      $v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }$

         $v= 1.76 *10^{14} m/s$