An insulating sphere of radius a, centered at the origin, has a uniform volume charge density p. A spherical cavity is excised from the inside of the sphere.The cavity has radius a4 and is centered at position h→, where |h→|&lt;34a, so that the entire cavity is contained within thelarger sphere. Find the electric field inside the cavity. Express your answer as a vector interms of any or all of ρ, ϵ0, r→ and h→

Question

An insulating sphere of radius a, centered at the origin, has a uniform volume charge density p.  A spherical cavity is excised from the inside of the sphere.The cavity has radius a4 and is centered at position h→, where |h→|&amp;lt;34a, so that the entire cavity is contained within thelarger sphere. Find the electric field inside the cavity.  Express your answer as a vector interms of any or all of ρ, ϵ0, r→ and h→

Jordan Mitchell · Accepted Answer

Step 1a) The charge enclosed by the Gaussian surface in terms of volume charge density and the radius of the sphere is,q=ρVThe volume of the sphere is as follows:V=43πr3Here, R is the radius of the sphere.Substitute 43πr3 for V in equation q=ρV q=ρ(43πr3)=43πr3ρStep 2The area of the surface element is,A=4πr2Here, r is the distance from the charge to the field point.The electric field is calculated by using the Gauss’s law.E∫dA=qϵ0EA=qϵ0Substitute 4πr2 for A and 43πr3ρ for q.E(4πr2)=1ϵ0(43πr3ρ)Solve the above equation for electric field inside the sphere.E(4πr2)=1ϵ0(43πr3ρ)E=ρr→3ϵ0Step 3b) The radius of the cavity is r″=a4 and is centered at position h→. So, the distance from the charge to the field point for the remaining hollow sphere is r′=h→−a4The net electric field in the cavity is equal to the difference between the electric field due to the remaining hollow sphere and the electric field due to charge filling cavity. So, the net electric field inside the cavity is,E→(r→)=E′=E″=ρr→′3ϵ0−ρr→″3ϵ0=ρ3ϵ0(r→′−r→″)Substitute r′=h→−a4 and r″=a4E→(r→)=ρ3ϵ0((h→−a4)−a4)=ρh→3ϵ0

Pansdorfp6 · Accepted Answer

Step 1b) Net electric bield mside the CavityE≠t=E−E′=ρ(h→−a4)3ϵ0−ρ(a{)43ϵ0E→≠t=h→ρ3ϵ0

alenahelenash · Accepted Answer

Step 1 Using Gausss

An insulating sphere of radius a, centered at the origin, has a unifor

Answered question

Answer & Explanation

New Questions in Other