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Marrrta [24]
2 years ago
12

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

face. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these
Physics
1 answer:
muminat2 years ago
5 0

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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Volgvan

Answer:

1 m = 100 cm....so 2.5 m = (2.5 * 100) = 250 cm

a = 1st shelf

b = 2nd

c = 3rd

d = 4th

a + b + c + d = 250

b = 2a + 18

c = a - 12

d = a + 4

a + (2a + 18) + (a - 12) + (a + 4) = 250

5a + 10 = 250

5a = 250 - 10

5a = 240

a = 240/5

a = 48 cm <== 1st shelf

b = 2a + 18 = 2(48) + 18 = 114 cm <== 2nd shelf

c = a - 12 = 48 - 12 = 36 cm <== 3rd shelf

d = a + 4 = 48 + 4 = 52 cm <== 4th shelf

so 2nd shelf is 114 cm

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