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Gennadij [26K]
3 years ago
9

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

- thermore, suppose that the electric field at a point 10 cm from the center is measured to be 1610 N/C radially inward while the elec- tric field at a point 35 cm from the center is 120 N/C radially outward.
R 2R 3R
6 cm 18 cm
26 cm

Find the charge on the insulating sphere. Answer in units of C.
Find the net charge on the conducting sphere. Answer in units of C.
Find the total charge on the inner surface of the hollow conducting sphere.
Find the total charge on the outer surface of the hollow conducting sphere.
Physics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

-1.7908787542\times 10^{-9}\ C

3.4260289211\times 10^{-9}\ C

1.7908787542\times 10^{-9}\ C

1.6351501669\times 10^{-9}\ C

Explanation:

r = Radius

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electric field is given by

E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C

The charge is -1.7908787542\times 10^{-9}\ C

Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C

The charge is 3.4260289211\times 10^{-9}\ C

The charge inside will have the polarity changed

q=+1.7908787542\times 10^{-9}\ C

Outside the charge will be

3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C

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L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\

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