Charge is distributed throughout a spherical volume of radius R with a density p = ar², where a is aconstant (of unit C/m³, in case it matters). Determine the electric field due to the charge at points bothinside and outside the sphere, following the next few steps outlined.Hinta. Determine the total amount of charge in the sphere.Hint for finding total chargeQencl =(Answer in terms of given quantities, a, R, and physical constants ke and/or Eg. Use underscore ("_")for subscripts, and spell out Greek letters.)b. What is the electric field outside the sphere?E(r> R) =c. What is the electric field inside the sphere?Hint for E within sphere#3Question Help: Message instructorE(r < R) =Submit QuestionEс$4RG Search or type URL%5T^MacBook Pro6Y&7U*8900

Given that: Charge density, ρ=αr2a) Let us consider a spherical section of radius r and thickness dr Charge inside the spherical element,dq=ρ(dA)=αr2×4πr2×dr=4παr4drTotal charge:q=∫0Rdqq=∫0R4παr4drq=4παR55b) To determine the electric field E(r>R), let us consider a gaussian surface of radius r as shown. From Gauss law:∮E→.dA→=QencεoHere,Qenc=q=4παR55Hence:E(4πr2)=4παR55εoE(r>R)=αR55εor2c) To determine the electric field E(r<R), let us consider a gaussian surface of radius r as shown. From Gauss law:∮E→.dA→=QencεoCharged enclosed by the gaussian surface:Qenc=∫0rρ(4πr2)drQenc=∫0r(αr2)(4πr2)drQenc=4παr55Then:E×4πr2=4παr55εoE (r<R) =αr35εo

Answered: Charge is distributed throughout a…