Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book

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11.3 Solutions 519

11.73 Q = C

V = 5 ×10

−6

× 250 = 1.25 × 10

−3

For the parallel connection

C = C

+ C

= 5 ×10

−6

+ 20 × 10

−6

= 25 ×10

−6

The resulting voltage



1.25 × 10

−3

25 × 10

−6

= 50 V

11.74 The combination of 2 μF and 2 μF in parallel is equivalent to 4 μF. This in

series with 8 μF gives a combined capacitance of 8/3 μF; 12 μF and 6 μF

in series gives an equivalent capacitance of 4 μF. 4 μF with 4 μF in parallel

gives 8 μF which in series with 1 μF yields 8/9 μF.

Combination of 8/9 μF and 8/3 μF in parallel gives 32/9 μF.

Effective value of C with 32/9 μF in series gives

C +

= 1, by problem

∴ C = 1

μF

11.75 Combined capacitance C for the three capacitors in series:

C =

+ C

(4 × 3 × 2) × 10

−18

(4 × 3 + 3 × 2 +2 × 4) × 10

−12

= 0.923 ×10

−6

(a)

q = CV = 0.923 × 10

−6

× 260 = 240 × 10

−6

∴ q

= q

= 240 ×10

−6

(b)

240 × 10

−6

4 × 10

−6

= 60 V

240 × 10

−6

3 × 10

−6

= 80 V

240

= 120 V

(c)

× 4 × 10

−6

× (60)

= 0.0072 J

× 3 × 10

−6

× (80)

= 0.0096 J

× 2 × 10

−6

× (120)

= 0.0144 J

520 11 Electrostatics

11.76 Charge on the ﬁrst capacitor

= C

V = 1 ×10

−6

× 12 = 12 × 10

−6

Charge on the second capacitor

= C

V = 2 ×10

−6

× 12 = 24 × 10

−6

Capacitance for the parallel combination

(a)

C = C

+ C

= (1 +2) ×10

−6

= 3 ×10

−6

q = q

= (12 +24) × 10

−6

= 36 ×10

−6

V =

36 × 10

−6

3 × 10

−6

= 12 V

(b)



= q

−q

= (24 −12) × 10

−6

= 12 ×10

−6

V =



12 × 10

−6

3 × 10

−6

= 4V

11.77 (a) If the positive end of a capacitor of capacitance C

, charged to potential

difference V

, is connected in parallel to the positive end of the capacitor

of capacitance C

charged to potential difference V

, then conservation

of charge gives the equation

+ C

)V = C

+ C

(1)

∴ V =

+ C

(common potential) (2)

The energy loss

W =

−

+ C

) V

− V

)

(3)

where we have used (2).

(b) If the positive end is joined to the negative end, the common potential

difference will be

V =

− C

+ C

(4)

and the energy loss will be

W =

+ V

)

(5)

11.3 Solutions 521

11.78

(a) Battery remains connected

(i) V



= V ; potential remains unchanged.

(ii) E



= E; electric ﬁeld is unchanged.

(iii) q



= Kq; charge is increased by a factor K . The additional charge

(K − 1)q is moved from the negative to the positive plate by the

battery, as the dielectric slab is inserted.

(iv) C



= KC; capacitance is increased by a factor K .

(v) U



KqV = KU

Energy is increased by a factor K .

(b) The battery is disconnected

(i) V



; potential is decreased by a factor K

(ii) E



; electric ﬁeld is decreased by a factor K . Both (i) and (ii)

follow from the fact that q



= q so that C



= CV and V



. Same reasoning holds good for E



(iii) q



= q; charge remains unchanged as there is no path for charge

transfer.

(iv) C



= KC; capacitance is increased by a factor K .

(v) U



The energy is lowered by a factor K .

11.79

(a) The battery remains connected

(i) V



= V ; potential remains unchanged.

(ii) E



< E; the electric ﬁeld is decreased since E = V/d, and V is

constant.

(iii) C



< C; capacitance is reduced since C ∝ 1/d.

(iv) q



< q; the charge is reduced since q = CV, with C decreas-

ing and V remaining constant. Some charge is transferred from the

capacitor to the charging battery.

(v) U



< U; the energy is decreased since U =

qV, with q decreas-

ing and V remaining constant.

(b) Battery is disconnected

(i) V



> V , the potential increases because q = CV, with C decreas-

ing and q remaining constant.

(ii) E



= E, the electric ﬁeld is constant because q = CV =

AE, with q remaining constant.

(iii) C



< C; the capacitance is decreased since C ∝ 1/d.

522 11 Electrostatics

(iv) q



= q; the charge remains constant.

(v) U



> U; energy increases because U =

qV, with V increasing

and q remaining constant.

11.80 As the plates carry equal but opposite charges, the force of attraction, which

is conservative, is given by

F =−

But U =

For the parallel plate capacitor,

C =

where x is the distance of separation. Combining the above equations

F =−





=−

2ε

=−

where we have put x = d.

11.81

(a) As the drops are assumed to be incompressible, the volume does not

change.

π R

= n

πr

∴ R = n

1/3

(b)



= 4πε

R = 4πε

1/3

∴ C



= n

1/3



4πε

1/3

2/3

4πε

= n

2/3

(d) σ



4π R

4π r

2/3

= n

1/3

(e) U



nqn

2/3

V = n

5/3

11.3 Solutions 523

11.82 The electric ﬁeld for a cylindrical capacitor is

E =

2πε

(1)

where l is the length and r the radius. The energy density (energy/unit vol-

ume)

u =

8π

(2)

wherewehaveused(1).

The energy stored between the coaxial cylinders of length l and radii R and

a is

U =



udv =



u (2πrl) dr (3)

where dv = (2πrdr)l is the volume element. Using (2) in (3)

U =

4πε



4πε

Similarly, the energy stored between the coaxial cylinders of radii b and a is

4πε

∴

ln (R/a)

ln (b/a)

Set

∴

ln (R/a)

ln (b/a)

→ ln

= 2ln

= ln

→ R =

√

11.83 The charge on C

= C

= 3 ×10

−6

× 4000 = 0.012 C

524 11 Electrostatics

The charge on C

= C

= 6 ×10

−6

× 3000 = 0.018 C

The combined capacitance

C =

+ C

3 × 6

3 + 6

× 10

−6

= 2 ×10

−6

Take the lower charge to ﬁnd the maximum voltage V :

V =

0.012

2 × 10

−6

= 6000 V

11.84 If the dielectric is present, Gauss’ law gives



E · ds = q − q



(1)

where −q



is the induced surface charge, q is the free charge and K is the

dielectric constant. Construct a Gaussian surface in the form of a coaxial

cylinder of radius r and length l, closed by end caps. Applying (1),

E (2πrl) =

or E =

2πε

rlk

(2)

In (1) the integral is contributed only by the curved surface and not the end

caps. The potential difference between the central rod and the surrounding

tube is given by

V =−



E · dr =



E dr =



2πε

The capacitance is given by

C =

2πε

ln (b/a)

11.85 The ﬁeld at point P is caused entirely by the charge Q on the inner sphere,

Fig. 11.45, and has the value

E =

4πε

11.3 Solutions 525

The potential difference between the two spheres is given by

V =−



E · dr =−

4πε



Q (b −a)

4πε

whence C =

4πε

b − a

Fig. 11.45

11.86 By prob. (11.85)

C =

4πε

b − a

(1)

Let b = a + where  is a small quantity. Then (1) can be written as

C =

4πε

a(a +  a)

a



4πε

a

(2)

Now the surface area A = 4π a

and a = d, the distance between the

surfaces, so that

C 

(parallel plate capacitor)

11.87

(i) q = q

−t/RC

100

∴ t = RC ln

= 1 ×10

× 10 × 10

−6

× 0.1056 = 1.056 s

526 11 Electrostatics

(ii) At one time constant t = RC

V = V

−1

120

2.718

= 44.15 V

U =

× 10 × 10

−6

× (44.15)

= 0.097 J

(iii) H = i

Rt =

t = V

C = (44.15)

× 10 × 10

−6

= 0.0195 J

11.88 Equilibrium energy U

Energy at time t

U =



1 − e

−t/RC



= U



1 − e

−t/RC





1 − e

−t/RC



Solving t = 1.228RC .

Thus after 1.228 time constants the energy stored in the capacitor will reach

half of its equilibrium value.

11.89 Let the capacitor be divided into differential strips which are practically par-

allel. Consider a strip at distance x of length a perpendicular to the plane

of paper and of width dx in the plane of paper, t he area of the strip being

dA = adx,Fig.11.46. At the distance x, the separation of the plates is seen

to be t = D + xθ. The capacitance due to the differential strip facing each

plate is

dC =

adx

D + x θ

Fig. 11.46

11.3 Solutions 527

The capacitance is given by

C =



dC =



a dx

D + x θ

= ε



D + x θ





1 +

x θ



−1





1 −

xθ

+···



dx 



x −







1 −

aθ



Note that for θ = 0, capacitance reduces to that for the parallel plate capaci-

tor.

11.90

(a) The equivalent capacitance of C

and C

in parallel is C

= 8 + 4 =

12 μF.

The combined capacitance of C

and C

in series is C =

+ C

3 × 12

3 + 12

= 2.4 μF.

Applied charge q = CV = 2.4 × 100 = 240 μC. Therefore charge on

will be q

= 240 μC. PD across C

will be V

240

= 80 V.

The PD across C

and C

will be equal.

= V

= (V − V

) = (100 − 80) = 20 V

(b) Now

(∵ V

= V

)

∴ q

= 2q

Also q

= 240

∴ q

= 160 μC and q

= 80 μC

(c)

× 8 × 10

−6

× 20

= 0.0016 J

× 4 × 10

−6

× 20

= 0.0008 J

528 11 Electrostatics

× 3 × 10

−6

× 80

= 0.0096 J

Note that U

= 0.012 J = U =

11.91

(a) The combination of C

and C

in series yields C

8 × 4

8 + 4

= 2.667 ×

−6

C = C

= (2.667 +3.0) × 10

−6

= 5.667 ×10

−6

q = CV = 5.667 × 10

−6

× 100 = 5.667 × 10

−4

= C

V = 3 ×10

−6

× 100 = 3 × 10

−4

= q

= q − q

= (5.667 −3.0) ×10

−4

= 2.667 ×10

−4

(b) V

= 100 V

2.667 × 10

−4

8 × 10

−6

= 33.33 V

2.667 × 10

−4

4 × 10

−6

= 66.66 V

(c)

× 8 × 10

−6

× (33.33)

= 0.00444 J

× 4 × 10

−6

× (66.66)

= 0.00889 J

× 3 × 10

−6

× 100

= 0.015 J

Note that U

= U =

, as it should.

11.92 Let the effective capacitance between points a and b be C. Apply a potential

difference V between a and b and let C be charged to q,Fig.11.47.

Let the charge across C

and C

be q

and q

, respectively; the charges across

various capacitors are shown in Fig. 11.47.

Fig. 11.47