Q66P Two isolated, concentric, condu... [FREE SOLUTION]

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Chapter 24: Q66P (page 714)

Two isolated, concentric, conducting spherical shells have radii $R_{1} = + 0.500 m$ and $R_{2} = 1.00 m$ , uniform charges $q_{1} = + 2.00 渭颁$ and $q_{2} = + 1.00 渭颁$ , and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) $r = 4.00 m$ , (b) $r = 0.700 m$ , and (c) $r = 0.200 m$ ? With $V = 0$ at infinity, what is $V$ at (d)localid="1663936396588" $r = 4.00 m$ , (e)localid="1663936405423" $r = 1.00 m$ , (f) localid="1663936413104" $r = 0.700 m$ , (g)localid="1663936424825" $r = 0.500 m$ , (h)localid="1663936434140" $r = 0.200 m$ , and (i)localid="1663936441939" $r = 0$ ? ( j) Sketch localid="1663936449400" $E (r)$ and localid="1663936460268" $V (r)$ .

Short Answer

Expert verified

a) The magnitude of electric field is $E = 1.687 脳$ $10^{3} N / C$ .

b) The magnitude of electric field is $E = 3.67 脳$ $10^{4} N / C$ .

c) The magnitude of electric field is $E = 0$ .

d) The value of electric potential is $V (r) = 6.75 kV$ .

e) The value of electric potential is $V (r) = 27 kV$ .

f) The value of electric potential is $V (r) = 34.7 kV$ .

g) The value of electric potential is $V (r) = 45 kV$ .

h) The value of potential is $V = 4.5 脳 10^{3} V$ .

g) The value of potential is $V = 45 脳 10^{3} V$ .

Step by step solution

Given data:

For the first spherical shell:

The radius, $R_{1} = 0.5 m$

The charge, $q_{1} = + 2.0 渭 C = + 2.0 脳 10^{- 6} C$

For the second spherical shell:

The radius, $R_{2} = 1.00 m$

The charge, $q_{2} = 1.0 渭 C = 1.0 脳 10^{- 6} C$

Understanding the concept:

The electric potential is define by using following formula.

$V = \frac{1}{4 蟺蔚_{o}} (\frac{q_{1}}{R_{1}} + \frac{q_{2}}{R_{2}}) = k (\frac{q_{1}}{R_{1}} + \frac{q_{2}}{R_{2}})$

Here, $V$ is the electric potential, $蔚_{0}$ is the permittivity of free space, $q_{1}$ and $q_{2}$ are the charges, and $R_{1}$ and $R_{2}$ are the distance, $k$ is the Coulomb鈥檚 constant having a value

(a) Calculate the magnitude of the electric field E at radial distance r=4.00 m:

At a radial distance, $r = 4.0 m$

$r > R_{2} > R_{1}$

The electric field is,

$E = k \frac{q_{1} + q_{2}}{r^{2}} = 9.0 脳 10^{9} 脳 \frac{(2.0 脳 10^{- 6} + 1.0 脳 10^{- 6})}{{(4.0)}^{2}}$

$= 1.687 脳 10^{3} N / C$

(b) Calculate the magnitude of the electric field E at radial distance r=7.00 m:

For radius, $r = 0.7 m$

R_{2} > r > R_{1}

The electric field is,

$E = k \frac{q_{1}}{r^{2}} = 9.0 脳 10^{9} 脳 \frac{2.0 脳 10^{- 6}}{{(0.7)}^{2}} = 3.67 脳 10^{4} N / C$

Step 5: (c) Calculate the magnitude of the electric field E at radial distance r=0.200 m:

If $R_{2} > R_{1} > r$

The electric field, $E = 0$

Because no charge lies inside the shell.

For $r = 0.2 m$ , $R_{2} > R_{1} > r$

Therefore. the electric field, $E = 0$

(d) With V = 0 at infinity, calculate V at r=4.00 m :

At infinity the potential, $V = 0$

If $r = 4.0 m$ , $r > R_{2} > R_{1}$

Thus, the electric potential is,

$V (r) = k (\frac{q_{1} + q_{2}}{r}) = 9.0 脳 10^{9} 脳 (\frac{1 脳 10^{- 6} + 2 脳 10^{- 6}}{4.0}) = 6.75 脳 10^{3} V = 6.75 kV$

(e) With V at infinity, calculate V at r=1.00 m :

For $r = 1.00 m$ ; $R_{2} > r > R_{1}$

The electric potential is,

$V (r) = k (\frac{q_{1}}{r} + \frac{q_{2}}{R_{2}}) = 9.0 脳 10^{9} (\frac{2.0 脳 10^{- 6}}{1.0 m} + \frac{1 脳 10^{- 6}}{1.0 m}) = 27 脳 10^{3} V = 27 kV$

(f) With V = 0 at infinity, calculate V at r=0.700 m :

For $r = 0.7 m$ ; $R_{2} > r > R_{1}$

The electric potential is,

$V (r) = k (\frac{q_{1}}{r} + \frac{q_{2}}{R_{2}}) = 9.0 脳 10^{9} (\frac{2.0 脳 10^{6}}{0.7} + \frac{1 脳 10^{- 6}}{1.0}) = 34.7 脳 10^{3} V = 34.7 kV$

(g) With V=0 at infinity, calculate V at r=0.500 m :

For $r = 0.5 m$ ; $R_{2} > r > R_{1}$

The electric potential is,

$V (r) = k (\frac{q_{1}}{r} + \frac{q_{2}}{R_{2}}) = 9.0 脳 10^{9} (\frac{2 脳 10^{- 6}}{0.5} + \frac{1 脳 10^{- 6}}{1.0})$

$V (r) = 45 脳 10^{3} V = 45 kV$

(h) With V=0 at infinity, calculate V at r=0.200 m :

For $r = 0.2 m$ ; $R_{2} > R_{1} > r$

$V (r) = k (\frac{q_{1}}{R_{1}} + \frac{q_{2}}{R_{2}}) = 9.0 脳 10^{9} (\frac{2 脳 10^{- 6}}{0.5} + \frac{1 脳 10^{- 6}}{1.0}) = 4.5 脳 10^{3} V$

(i) With V = 0 at infinity, calculate V at r=0 m :

For $r = 0$ ; $R_{2} > R_{1} > r$

$V (r) = k (\frac{q_{1}}{R_{1}} + \frac{q_{2}}{R_{2}}) = 9.0 脳 10^{9} (\frac{2 脳 10^{- 6}}{0.5} + \frac{1 脳 10^{- 6}}{1.0}) = 45 脳 10^{3} V$

(j) Sketch E(r)and V(r) :

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Short Answer

Step by step solution

Given data:

Understanding the concept:

(a) Calculate the magnitude of the electric field E at radial distance r=4.00 m:

(b) Calculate the magnitude of the electric field E at radial distance r=7.00 m:

Step 5: (c) Calculate the magnitude of the electric field E at radial distance r=0.200 m:

(d) With V = 0 at infinity, calculate V at r=4.00 m :

(e) With V at infinity, calculate V at r=1.00 m :

(f) With V = 0 at infinity, calculate V at r=0.700 m :

(g) With V=0 at infinity, calculate V at r=0.500 m :

(h) With V=0 at infinity, calculate V at r=0.200 m :

(i) With V = 0 at infinity, calculate V at r=0 m :

(j) Sketch E(r)and V(r) :

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