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Ksju [112]
3 years ago
12

13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distanc

e and the direction. We will investigate in detail just two directions. With charges available in the simulation how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right
Physics
1 answer:
Gennadij [26K]3 years ago
5 0

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of  charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

               P→

_{-\theta } (-) ---------------->(+) _{+\theta }

               d

so let distance between the dipoles be d

∴ P = d\Theta

Let \Theta_{1} = 1 nC

so

P = d\Theta

1 × 10⁻⁹ =  1 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let \Theta_{2} = 2 nC

so

P = d\Theta

1 × 10⁻⁹ =  2 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)  

d = 0.5 m

Also Let \Theta_{3} = 3 nC

so

P = d\Theta

1 × 10⁻⁹ =  3 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge                 distance

1 nC                        1.00 m      

2 nC                       0.50 m

3 nc                        0.33 m

4 nC                       0.25 m

5 nC                       0.20 m      

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Answer:

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Explanation:

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A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]            ----------------------    (1)

in which R₁ is the resistance at temperature T₁

R_{T(2) is the resistance at temperature T₂

Given that V= IR

R = \frac{V}{I}

Therefore, the resistance at temperature 28°C is;

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R_{T(2) = \frac{120V}{1.23A}

= 97.56Ω

From (1) above;

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]      

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)

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T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

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P = I²₁R₂₈

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P = 163.209 W

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P= I^2_2R_{T^0C

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

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3 years ago
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