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

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.