1.) A capacitor consists of two closely spaced metal conductors of large area, s

Question

1.) A capacitor consists of two closely spaced metal conductors of large area,
separated by a thin insulating foil. It has an electrical capacity of
3800 uF and is charged to a potential difference of 75.0 V. Calculate the amount of energy stored in the capacitor.

2.)Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 9.85 J.

3.)If the two plates of the capacitor have their separation increased by a factor of 4 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Explanation / Answer

Remember and apply for parallel plate caapcitor, these as formulas

1. Capacitance C = K eoA/d

where K is dieelctric constant   ( for air K = 1)

eo is permnittivity constant = 8.85 e-12

A is area = pi^2   ( for circular plates)    and   l* b   for rectanular plates

and

d is the distance between the plates

2.Charge Q = CV     where V is potrntial dif

3. energy stored U = 0.5 QV      or 0.5 CV^2    or 0.5 Q^2/C

4. electric field between the plates is E = Q/eo A    ( also given by E = V/d)

so here now we apply

1. energy stored U = 0.5 * 3800e-6 * 75*75

U = 10.68 Joules -------<<<<<<<<<<<<<<<<<answer to part 1

----------------------------------------------

if u = 9.85 Joules

then

Q^2 = 2UC    = 2*9.85* 3800e-6

Q ^2 =0.0736

Q = 0.273 Coulombs -----------<<<<<<<<<<<<<<-answer to part B

-----------------------

part C:

if dnew = d*4

then

C = 3800e-6*4   = 15200 uF

so

enenrgy New = 0.5 * 15200 e-6 * 75*75

Unew = 42.75 Joules

so change ofe nergy U = 42.75 - 9.85

U = 32.9 Joules   or 33 J -------<<<<<<<<<<<<<<<<<<<is the answer

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