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Leno4ka [110]
3 years ago
9

Three capacitors having capacitances of 8.40, 8.40, and 4.20μF, respectively, are connected in series across a 36.0-V potential

difference. (a) What is the charge on the 4.20−μF capacitor? (b) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination?
Physics
1 answer:
Vlad [161]3 years ago
3 0

Answer:

Therefore,

The charge on the 4.20μF capacitor is 75.6μC.

The voltage across each capacitor in the parallel combination is 10.8 Volt.

Explanation:

Given:

Three capacitors having capacitance,

C_{1}=8.4\mu F\\C_{2}=8.40\mu F\\C_{3}=4.20\mu F\\

are connected in series across a 36.0-V potential difference,

V= 36 V

To Find:

Qs = ? charge on the 4.20−μF

V = ? voltage across each capacitor in the parallel combination

Solution:

For Capacitor Series Combination we have,

\dfrac{1}{C_{s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}

Substituting the values we get

\dfrac{1}{C_{s}}=\dfrac{1}{8.4}+\dfrac{1}{8.4}+\dfrac{1}{4.2}\\\\C_{s}=2.1\mu F

As Capacitors are connected in series they have the same charge on each plates, and is given by

Q= C_{s}\times V

Substituting the values we get

Q= 2.1\mu F\times 36=75.6\mu C\\Q=75.6\mu C

Now when the plates are connected in Parallel we will have,

C_{p}=C_{1}+C_{2}+C_{3}

Substituting the values we get

C_{p}=8.4+8.4+4.2=21\mu F

As Capacitors are connected in Parallel they have the same Voltage on each plates, and is given by

V= \dfrac{Q_{1}+Q_{2}+Q_{3}}{C_{p}}

Substituting the values we get

V=\dfrac{75.6+75.6+75.6}{21}=10.8\ V

Therefore,

The charge on the 4.20μF capacitor is 75.6μC.

The voltage across each capacitor in the parallel combination is 10.8 Volt.

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<u>Electrostatic Force</u>

Two point-charges q_1 and q_2 separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

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\displaystyle F_1=\frac{k\ Q^2}{d^2}

The force between charges separated 2d is

\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}

\displaystyle F_2=\frac{k\ Q^2}{4d^2}

And the force between the charges A and D is

\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}

\displaystyle F_3=\frac{k\ Q^2}{9d^2}

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_A=-\frac{41}{36}F_1

Charge B. It receives force to the right from A and D and to the left from C

\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}

\displaystyle F_B=\frac{1}{4}F_1

Charge C. It receives forces to the right from all charges.

\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}

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\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}

\displaystyle F_D=-\frac{49}{36}F_1

Comparing the magnitudes of each force is just a matter of computing the fractions

\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36

a.

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b.

The ratio of the greatest to the smallest net force is

\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9

The greatest force is 9 times the smallest net force

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