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Katena32 [7]
3 years ago
11

What was the purpose of the little Albert experiment?

Physics
1 answer:
Murljashka [212]3 years ago
3 0

Answer:

to condition a phobia in an emotionally stable child

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Answer:

a) 4.94e9 J  b) 1.07e10 J

Explanation:

The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

U =\frac{1}{2}*C*V^{2}

a) For the original capacitor, we can find directly U as follows:

U =\frac{1}{2}*1.81F*(73.9e3)^{2} V2 = 4.94e9 J  

U = 4.94*10⁹ J

b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.

As the capacitance is proportional to the dielectric constant, we can write the following proportion:

ε₂ / ε₁ = \frac{943}{435}= \frac{C2}{C1} =\frac{Cx}{1.81F}

Cx =\frac{1.81F*943}{435} = 3.92 F

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

U =\frac{1}{2}*3.92F*(73.9e3)^{2} V2 = 1.07e10 J

⇒ U = 1.07*10¹⁰ J

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