The charge distribution in this problem is spherically symmetric, so you can solve it with Gauss's law. The charge within a sphere of radius ~ includes the proton charge + @ plus the portion of the electron charge distribution that lies within the sphere. The electron charge distribution is not uniform, so the charge enclosed within a sphere of radius + is not simply the charge density multiplied by the volume say of the sphere. Instead, you'll have to do an integral. Part A Consider a thin spherical shell centered on the proton, with radius y, and infinitesimal thickness dry. Since the shell is so thin, every point within the shell is at essentially the same radius from the proton. Hence the amount of electron charge within this shell is equal to the electron charge density p(r) at this radius multiplied by the volume dV of the shell. What is dl in terms of ri? Express your answer in terms of the variables ~, and dri. > View Available Hint(s) dy = 4mri dri Submit Previous Answers CorrectThe total electron charge within a radius ~ equals the integral of p(m ) @V from = 0 tom, =r. Set up this integral (but don't solve yet), and use it to write an expression for the total charge (including the proton) within a sphere of radius r. View Available Hint(s) O The expression for the electron's charge disturbed in the space is -q()sphere = 0 ( ) e-2 1/304mm; dri, the expression for the total charge is Cina = @proton - 9(") sphere. O The expression for the electron's charge disturbed in the space is -q( r ) sphere = Jo - ( e-271/00 Amri dy1, the expression for the total charge is @ind = @proton + 9(")sphere. The expression for the electron's charge disturbed in the space is -9(7) sphere = Jo - Tan e-271/20 4my) dry, the expression for the total charge is Gina = @proton - 9(r)sphere. The expression for the electron's charge disturbed in the space is -q(")sphere - fo - ( mars ) e-271/00 4mri dry , the expression for the total charge is Qina = 0. Submit Previous Answers CorrectReview | Constants A hydrogen atom is made up of a proton of charge Submit Previous Answers +Q = -1.60 x 10-19 C and an electron of charge -Q = -1.60 x 10- C. The proton may be regarded as a point charge at r = 0, the center of the atom. Correct The motion of the electron causes its charge to be "smeared out" into a spherical distribution around the proton (Figure 1), so that the electron is equivalent to a charge per unit volume of EXECUTE P(T) = -(Q/map )e-2r/0, where an = 5.29 x 10 m is called the Bohr radius. (a) Find the total amount of the hydrogen atom's charge Part C that is enclosed within a sphere with radius r centered on the proton. (b) Find the electric field (magnitude and direction) caused by the charge of the Integrate the expression and solve the linear system from previous part to find the total hydrogen atom as a function of r. c) Make a graph as charge Q(r)enc] within radius r. a function of r of the ratio of the electric-field magnitude E to the magnitude of the field due to the Express your answer in terms of the variables Q, r, and ag. proton alone. View Available Hint(s) Q()enel = Qe 2( 20) + 2+1) Submit Previous Answers All attempts used; correct answer displayed Part D Use Gauss's law and your previous result to find the electric field at a distance r from proton; also find the electric field created by a proton itself. Express your answer in terms of the variables r, Q, ao, k (the electric force constant), and any other appropriate constants separated by comma. View Available Hint(s) IVO AEd Figure E(r)space, E(r )proton = Submit Proton: point charge + Part E Complete previous part(s) EVALUATE Electron: charge -Q "smeared out" in a spherical distribution Part F Complete previous part(s)