Question

Problem 2: a conducting sphere A conducting sphere has a positive net charge Q and radius R. (Note: since the sphere is conducting all the charge is distributed on its surface.) a) By reflecting on the symmetry of the charge distribution of the system, determine what the E-field lines look like outside the sphere for any r > R. Describe the E-field in words and with a simple sketch. Make sure to also show the direction of the E-field lines. b) Consider a point P, shown in the figure below, at a distance rp from the center of the sphere. Your goal for this part is to properly use Gauss' law to calculate the electric field at point P. Choose an appropriate Gaussian surface and clearly draw it on a sketch of the wire. Show your calculation for the electric flux through the Gaussian surface, as well as for the charge enclosed. What at is the electric field at point P? Give both the magnitude and direction (give the magnitude in terms of the given quantities: Q, R, P) lp c) Write an expression for the magnitude of the E field for any radius r > HR. d Similarly to part b) use Gauss' law to determine the magnitude of the electric field for any radius r inside the sphere, therefore for 0<r< R e) Make a sketch of the magnitude of E(r) for any r20. Clearly show on your graph the behavior of E(r) for both regions: 0rR and r> R.

Answer 1

Electric field line always point away from the positive charge, hence the vectors are pointing in outwards direction and this electric field is symmetric as the charge is located all over the surface of sphere. Consider a sphere of radius r_p rightarrow This is the Gaussian surface. \r\n According to Gauss's law- Contourintegral = Contourintegral vector E.middot d vector s = Q_enclosed/E_0 E_x y pi r p^2 = Q_closed/E_0 E = Q/E_0 times 4 pi r^2_p and is directed away from the point p. as electric field due to a positive charge is away from it. For any radius r greaterthan R. Consider any Gaussian surface of radius r greaterthan R Conotourintegral = Contourintegral vector E.midot d vector s = Q/E_0 E times y pi r^2 = Q/E_0 \r\n E = Q/4 pi e_0 r^2 for a_r < R Contourintegral = Contourintegral vector E middot d vector s = Q_enclosed/e_0 As whole of the charge is located at the surface