A spherical conductor with a 0.193 m radius is initially uncharged. How many electrons should be removed from the sphere in order for it to have an electrical potential of 7.10 kV at the surface?

Respuesta :

Answer:

The number of electrons removed is  [tex]N = 9.513 *10^{11}[/tex]

Explanation:

From hr question we are told that

    The radius is  [tex]r = 0.193 \ m[/tex]

   The required electrical potential is [tex]V = 7.10 \ kV = 7.10 *10^{3} V[/tex]

The total charge on the sphere  is mathematically evaluated as

           [tex]Q = \frac{Vr}{k}[/tex]

where k is the coulombs constant with value [tex]k =9)*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.[/tex]

substituting value

          [tex]Q = \frac{7.10 *10^3 * 0.193}{9*10^9}[/tex]

          [tex]Q = 1.522*10^{-7} C[/tex]

The number of electron removed is mathematically evaluated as

        [tex]N = \frac{Q}{e}[/tex]

Where e is  the charge on one electron with value [tex]e = 1.6 *10^{-19} \ C[/tex]

  substituting values  

      [tex]N = \frac{1.522 *10^{-7}}{1.60*10^{-19}}[/tex]

      [tex]N = 9.513 *10^{11}[/tex]