Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

12. -10 points SerPSE9 25.P.059 Notes Ask Your Teache + The electricpotential im

ID: 1582886 • Letter: 1

Question

12. -10 points SerPSE9 25.P.059 Notes Ask Your Teache + The electricpotential immediately outside a charged conducting sphere is 240 V and 10.0 cm farther from the center of the sphere the potential is 120 v (a) Determine the radius of the sphere. cm ) Determine the charge on the sphere nc The electric potential immediately outside another charged conducting sphere is 260 V, and 10.0 cm farther from the center the magnitude of the electric field is 410 V/m. (c) Determine all possible values for the radius of the sphere. (Enter your answers from smallest to largest. If only one value exists, enter "NONE in the second answer blank.) cm cm (d) Determine the charge on the sphere for each value of r. (If only one value exists, enter "NONE" in the second answer blank.)

Explanation / Answer

electric potential V1 = 240 V

electric potential V2 = 120 V

electric potential at the surface of the sphere is

                       V1 = kq/ R   

                    240 = kq/ R

                       kq = 240*R   ........... (1)

potential at (R + 10) from the center of the sphere is

                   V2 = kq/(R + 10)

                  120 = kq/(R + 10)

                     kq = 120(R + 10)    ........... (2)

compare eq (1) and (2), we get

                  240*R  = 120(R + 10)

radius of the sphere is

                      R = 10 cm = 0.10 m

.......................................................................................

substitute the value R = 10 cm in eq (1), we get

                     kq = 240*0.1

                    (9*109)(q) = 24

                  q = 2.67*10-9 C = 2.67 nC

charge on sphere, q = 2.67 nC

electric potential at the surface of the sphere is

                       V1 = kq/ R   

                    260 = kq/ R

                       kq = 260*R ................... (3)

electric field at (R + 10) from the center of the sphere is

         E = kq/(R + 10)2

     410 = kq/(R + 10)2

       kq = 410(R + 10)2 ..................... (4)

compare eq (3) and (4), we get

        260*R = 410(R + 10)2

        260*R = 410(R + 0.1 m)2

         410 R2 - 178 R + 4.1 = 0

solve the above quadratic equation , we get

      R = 0.41 m or R = 0.0244 m

       R = 41 cm or R = 2.44 cm

radius of the sphere is,

          R = 41 cm (or) R = 2.44 cm

if R = 41 cm ,
substitute the value R = 41 cm in eq (3), we get
charge on the sphere q = 11.84 nC
if R = 2.44 cm ,
substitute the value R = 2.44 cm in eq (3), we get

charge on the sphere q = 1.11 nC