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Sergio039 [100]

3 years ago

5

A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou

ter radius 5.00 cm is concentric with the solid sphere and has a charge of −2.13 μC. Find the electric field at the following radii from the center of this charge configuration. (a) r = 1.00 cm (b) r = 3.00 cm (c) r = 4.50 cm (d) r = 7.00 cm

1 answer:

madreJ [45]3 years ago

5 0

Answer:

Part a)

E = 0

Part b)

$E = 6.77 \times 10^7 N/C$

Part c)

Electric field inside the conductor is again zero

$E = 0$

Part d)

$E = 8.52 \times 10^6 N/C$

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

$E = \frac{kq}{r^2}$

$E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}$

$E = 6.77 \times 10^7 N/C$

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

$E = 0$

Part d)

Now at r = 7 cm

again by Gauss law

$E = \frac{kQ}{r^2}$

$E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}$

$E = 8.52 \times 10^6 N/C$

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Which statement best explains water’s ability to dissolve covalent compounds? brainly.com/question/8223274
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