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2 years ago

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A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charg

es the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?

1 answer:

pogonyaev2 years ago

4 0

Answer:

The answer is "Option D".

Explanation:

Please find the complete question in the attached file.

As plate separation increased to 2d the capacitance get halred but the change remain same

$\therefore V=\frac{Q}{C}$

The voltage doubles are now electric field remain same because both the distance and voltage get doubled.

$\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\$

So,

$energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E$

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Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

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Part a)

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E_net = E_2 + E_4

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r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

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E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

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Part b)

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Hence,

E_net = 2*E_part(a)*cos(Q)

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Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

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E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

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