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3 years ago

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What is the beta of a 3-stock portfolio including 25% of stock A with a beta of 0.90, 40% of stock B with a beta of 1.05, and 35

% of stock C with a beta of 1.73

1 answer:

natka813 [3]3 years ago

5 0

Answer:

1.25

Explanation:

Calculation for What is the beta of a 3-stock portfolio

Portfolio beta = (.25 *0.9) + (.4 *1.05) + (.35 *1.73)

Portfolio beta = .225 + .42 + .606

Portfolio beta = 1.25

Therefore the beta of a 3-stock portfolio will be 1.25

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A point charge q1 = -9.6 μC is located at the center of a thick conducting spherical shell of inner radius a = 2.4 cm and outer

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Answer:

Part 1: The value of x component of electric field at point P is $-1.03 \times 10^7 \frac{ N}{ C}$

Part 2: The value of y component of electric field at point P is 0.

Part 3: The value of x component of electric field at point R is 0.

Part 4: The value of y component of electric field at point R is $-5.06 \times 10^8 \frac{ N}{ C}$ .

Part 5: The value of surface density at the outer edge of the shell is $-3.83 \times 10^{-4} C/m^2$ .

Part 6: None ,The field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of $r^2$ .

Part 7:The fields are equal as the charge on the outer shell does not affect the field on within the shell. ( $E_2=E_o$ )

Explanation:

Part 1

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$x= 8.4 cm = 0.084$ m is the location of P on x

So the value is given as

$E_x( P) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.084^2}\right) \\= -1.03 \times 10^7 \frac{ N}{ C}$

The value of x component of electric field at point P is $-1.03 \times 10^7 \frac{ N}{ C}$

Part 2

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$y = 0 cm = 0.0$ m is the location of P on y

So the value is given as

$E_y( P) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0$

The value of y component of electric field at point P is 0.

Part 3

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$x = 0 cm = 0.0$ m is the location of R on x

So the value is given as

$E_x( R) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0$

The value of x component of electric field at point R is 0.

Part 4

As

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

$y =1.2 cm = 0.012$ m is the location of R on y

So the value is given as

$E_y( R) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.012}\right) \\=-5.06 \times 10^{8} N/C$

The value of y component of electric field at point R is $-5.06 \times 10^8 \frac{ N}{ C}$ .

Part 5

As

$\sigma = \frac{ q_{enclosed}}{ 4 \pi r^2}$ is the Surface charge density

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

$a = 2.4 cm = 0.024$ m is inner radius of the conducting shell

$b = 4.1 cm = 0.041$ m is outer radius of the conducting shell

So the value is given as

$\sigma_b = \frac{ q_{enclosed}}{ 4 \pi b^2}\\\sigma_b = \frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{ 4 \pi 0.041^2}\\\sigma_b = -3.83 \times 10^{-4} C/m^2$

The value of surface density at the outer edge of the shell is $-3.83 \times 10^{-4} C/m^2$ .

Part 6

<em>None because the field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of </em> $r^2$ .

Part 7

<em>The fields are equal as the charge on the outer shell does not affect the field on within the shell. (</em> $E_2=E_o$ )

5 0

3 years ago