a) What is the radial component of the electric field E r at a point located at

Question

a) What is the radial component of the electric field Er at a point located at radius r = 2.40 cm, i.e. between the two conductors? Er is positive if it points outward, negative if it points inward.

b) What is Er at a point located at radius r = 3.60 cm, i.e. outside the outer shell?

c) What is the surface charge density, b, on the inner surface of the outer spherical conductor?

d) What is the surface charge density, c, on the outer surface of the outer spherical conductor?

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A solid metal sphere of radius a = 1.30 cm is surrounded by a concentric spherical metal shell of inner radius b = 2.70 cm and outer radius c = 3.20 cm. The inner sphere has a net charge of Q1 = 4.10 muC, and the outer spherical shell has a net charge of Q2 = - 7.90 muC. a) What is the radial component of the electric field Er at a point located at radius r = 2.40 cm, i.e. between the two conductors? Er is positive if it points outward, negative if it points inward. b) What is Er at a point located at radius r = 3.60 cm, i.e. outside the outer shell? c) What is the surface charge density, sigmab, on the inner surface of the outer spherical conductor? d) What is the surface charge density, sigmac, on the outer surface of the outer spherical conductor?

Explanation / Answer

a)The charge on the shell does not cause electric field to the point that have the radius of 2.4 cm so E=kq/r^2 = (8.99x10^9)(4.1x10^-6) / (2.4x10^-2)^2 = 6.406x10^ 7(V/m) Outward (positive) b.At r= 3.6cm, it is outside both the shell and the sphere so According to Gauss's law E*4pr^2=Q/e0 so E=kQ/r^2 =(8.99x10^9)(4.1x10^-6 - 7.9x10^-6) / (3.6x10^-2)^2 =-2.638x10^7(V/m) it is inward.(negative.) Since the outer sphere is conductor the whole charge resides on its surface The charge of the inner metal sphere induces an equal and opposite charge on the inner surface of the outer sphere Hence charge induced on the inner surface on the outer sphere = - 4.1x10^-6 C surface charge density s = -Q1 / 4pb^2 Here b = inner surface radius = 0.027 m sinner = -4.476x10^-4 C/m^2 Charge on the outer surface of the outer sphere q = Q2 + Q1 = - 7.9 + 4.1 = -3.8µC Surface charge density souter = q / 4pc^2 Here c = 0.032 m souter = -2.953x10^-4 C/m^2

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