For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 c

Question

For the configuration shown in the figure below, suppose a = 5.00 cm, b = 20.0 cm, and c = 25.0 cm. Furthermore, suppose the electric field at a point 16.5 cm from the center is measured to be 3.40 × 103 N/C radially inward and the electric field at a point 50.0 cm from the center is of magnitude 152 N/C and points radially outward. From this information, find the following. (Include the sign of the charge in your answer.) Insulator Conductor (a) the charge on the insulating sphere (b) the net charge on the hollow conducting sphere (c) the charge on the inner surface of the hollow conducting sphere (d) the charge on the outer surface of the hollow conducting sphere Need Help?Watch

Explanation / Answer

a]We know that, electric field E = kQenclosed/r^2

Charge on insulator = Qenclosed = E r^2/k

= - 3.4e3*0.165^2/9e9

= - 1.0285*10^-8 C answer

b] in second case, Qenclosed = E r^2/k = 152*0.5^2/9e9 = 4.2222*10^-9 C

charge on hollowe sphere = 4.2222*10^-9 - - 1.0285*10^-8 = 1.4507 *10^-8 C answer

c] charge on inner surface of hollow sphere = negative of insulator = 1.0285*10^-8 C answer

d] charge on outer surface of hollow sphere = 1.4507*10^-8 -1.0285*10^-8 C =   4.2222*10^-9 C answer

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