A thin nonconducting spherical shell of radius R1 carries a total charge q1 that

Question

A thin nonconducting spherical shell of radius R1 carries a total charge q1 that is uniformly distributed on its surface. A second, larger thin non-conducting spherical shell of radius R2 that is concentric with the first carries a charge q2 that is uniformly distributed on its surface.
(a) Use Gauss's law to find the electric field in the regions r < R1, R1 < r < R2, and r > R2. (Use any variable or symbol stated above along with the following as necessary: k.)

E (r < R1) =

E (R1 < r < R2) =

E (r > R2) =

What should the ratio of the charges q1/q2 and their relative signs be for the electric field to be zero for r > R2?

q1/q2 =

Please explain, thanks in advance

Explanation / Answer

Given the two spherical shells are non conductors Electric field at r < R1 constust a gaussian surface with this radius inside the sherical shell inside the sperical shell the charge enclosed   Q =0 Accrording to gauss law electric flux E (4r^2 )   = Q / _0   == > E = kQ /r^2 since Q =0 E (r<R1)   = 0 --------------------------------------------------- The electric field at a point E (R1 < r < R2 )   Electric field at R1 < r < R2 constust a gaussian surface with this radius Electric field at this point E = Eshell (R1)   + Eshell (R2) The electric field due to Eshell (R2)   = o , since the point is inside the outer shell Eshell (R1)    = kq1 / r^2 Electric field   (R1 < r < R2 )   = kq1 / r^2   + 0 =  kq1 / r^2 ---------------------------------------------------------- The electric field at a distace r > R2 outside   the outer spherical shell Electric field   E = Eshell (R1)   + Eshell (R2)   electric field due to inner shell Eshell (R1)    = kq1 /r^2   electric field due to outer shell Eshell (R2)   = kq2 /r^2   E (r > R2) =  kq1 /r^2   +  kq2 /r^2     --------------------------------- The ratio   of charges must be q1 / q2 = -1 then E (r > R2) = 0 q1 = + ve and q2 = -ve (OR) q1 = - ve and q2 = +ve either of two cases  E (r > R2) = 0   Electric field at R1 < r < R2 constust a gaussian surface with this radius Electric field at this point E = Eshell (R1)   + Eshell (R2) The electric field due to Eshell (R2)   = o , since the point is inside the outer shell Eshell (R1)    = kq1 / r^2 Electric field   (R1 < r < R2 )   = kq1 / r^2   + 0 =  kq1 / r^2 ---------------------------------------------------------- The electric field at a distace r > R2 outside   the outer spherical shell Electric field   E = Eshell (R1)   + Eshell (R2)   electric field due to inner shell Eshell (R1)    = kq1 /r^2   electric field due to outer shell Eshell (R2)   = kq2 /r^2   E (r > R2) =  kq1 /r^2   +  kq2 /r^2     --------------------------------- The ratio   of charges must be q1 / q2 = -1 then E (r > R2) = 0 q1 = + ve and q2 = -ve (OR) q1 = - ve and q2 = +ve either of two cases  E (r > R2) = 0  

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