A solid insulating sphere of radius a = 4.9 cm is fixed at the origin of a co-or

Question

A solid insulating sphere of radius a = 4.9 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density rho = -262.0 microC/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 12.6 cm, and outer radius c = 14.6 cm.

What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity.

What is V(c) - V(a), the potentital differnece between the outer surface of the conductor and the outer surface of the insulator?

Explanation / Answer

A solid insulating sphere of radius a = 4.9 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ? = -108 µC/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 12.8 cm, and outer radius c = 14.8 cm. 1) What is Ex(P), the x-component of the electric field at point P, located a distance d = 34 cm from the origin along the x-axis as shown? What is V(b), the electric potential at the inner surface of the conducting shell? Define the potential to be zero at infinity. What is V(a), the electric potential at the outer surface of the insulating sphere? Define the potential to be zero at infinity. 2. Relevant equations U12=k(q1q2/r12) k=9E9 3. The attempt at a solution I was able to find the answer to the first question using the equation i gave but when i tried that for 2 and 3 it didnt work and gave the message: "It looks like you have calculated the potential at the inner radius of the shell to be equal to the potential at r = c produced by the insulating sphere by itself. The conducting shell plays a role here. Go back to the definition of the potential to determine the answer." I don't really understand how the outside uncharged spherical conducting shell can have any effect when it is uncharged. I must be missing something here. Please give input. 1.Use Gauss's law.Imagine a spherical film of radius d=35cm.Now you have Ex x 4pd2=4/3xpa3?/e0 all are given except Ex.Find it 2.the insulator will induce a charge equal magnitude but opposite nature in the inner wall,causing the inner wall having no net charge.SO integretion of E.dl=0.That is V=0 3.total charge inside insulator=4/3xpa3=a,say.Potential=a/(4pe0a).everything given easy to find it out. 4.For charge conservation a charge +a will be induced in the outer wall.then net charge=a-a+a=a. potential at outer wall=a/(4pe0c).Now calculate V(c)-V(a) 5.Here the inner configuration is unaffected.V(a) will remain the same.Alternately,imagine a spherical film of radius a centred at origin.The charge inside it is same before and after addition of Q at the outer wall , from this fact E is unchanged causing an unchanged V. I haven't done this numerical calculations ,do it your self(convert CGS into SI first) Two identical square parallel metal plates each have an area of 500 cm2. They are separated by 1.50 cm. They are both initially uncharged. Now a charge of +1.50 nC is transferred from the plate on the left to the plate on the right and the charges then establish electrostatic equilibrium. (Neglect edge effects.) (a) What is the electric field between the plates at a distance of 0.25 cm from the plate on the right? (b) What is the electric field between the plates a distance of 1.00 cm from the plate on the left? (c) What is the electric field just to the left of the plate on the left? (d) What is the electric field just to the right of the plate on the right? Solution: The left plate has negative charge -1.50 nC and the right one has positive charge + 1.50 nC. The field due to each surface charge distribution can be calculated by the Gauss’ law applied to a cylinder that crosses each plane. For each plane: ES+ES = sS/e0 E = s/2e0 The field for the 2 plates sums inside the 2 plates and cancels outside. (a) and (b) the field is uniform inside the plates so in any point the value is: ! E = " #0 = Q/S #0 = 1.50 $10%9 8.85 $10%12 $ 500 $10%4 = 3.39kN /C (c) and (d) outside plates the fields due to the 2 plates cancel and the field is zero. Tipler 22.P.076 A nonconducting sphere 1.20 m in diameter with its center on the x axis at x = 4.00 m carries a uniform volume charge of density ? = +5.00 µC/m3. Surrounding the sphere is a spherical (thin) shell with a diameter of 2.40 m and a uniform surface charge density s = -1.50 µC/m2. Calculate the magnitude and direction of the electric field at the following locations. (a) x = 4.50 m, y = 0 (point 1 in the figure) (b) x = 4.00 m, y = 1.10 m (point 2) (c) x = 2.00 m, y = 3.00 m (point 3)

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