Ahollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37×10−6c/m2. a charge of −0.500μc is now introduced at the center of the cavity inside the sphere. (a) what is the new charge density on the outside of the sphere? (b) calculate the strength of the electric field just outside the sphere. (c) what is the electric flux through a spherical surface just inside the inner surface of the sphere?

a)5.73 *10^-6 C/m^2; b)648*10· N/C c)56.5 *10^3 Nm^2/C

Explanation: Considering the definition of surface charge density as:

σ  = Q/ area of sphere

so Q = σ *  4*π* R^2outer

The new Q at the outer radius is: Q initial -0.5 microC =5 micro C

The Electric with the new outer charge is equal to:

E= k*Q / R^2outer= 9 10^9* 5 microC/(0.25m)^2= 648*10^3 N/C

Finally the flux corresponding to the internal charge is equal to

By using Gauss law, the total flux is equal to net change divided ε0

so this is 0.5 microC/ ε0= 56.5 *10^3 Nm^2/C

trig 15.3 cos

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.