1)A capacitor of C = 500 mF is initially charged to a potential difference of 10.0 V and connected to a resistor R = 100 kW. Find:

a) The initial charge on the
capacitor

**b) **The time constant for
discharge through the resistor R.

**c) **The time for the initial
charge to decay to one half its initial value.

**d) **The energy stored in the
capacitor by this time.

__Solutions:__

**a)
**Q = CV
=**
**(500 mF) (10.0 V) = (5 x
10^{—4} F) (10.0 V) = 5 x 10^{—3} C

**b)
**t = CR = __
__ (5 x 10^{—4} F) (10 ** ^{5 }** W.) = 5 x
10

^{ 1}s

** c)
**Q =

*Q*

_{o}

*e*^{-(t/CR)}

Or: Q/ *Q _{o} *

**= ½ =**e

^{-(t/CR)}

** **2
= *e ^{(t/CR)}*

t = CR ln 2 = (50 s) (0.693) = 34.7 s

**d) **** Q**** **= *½ Q _{o} _{ }*=
2.5 x 10

^{—3}C

**W = **Q ^{2}
/ 2C =
(2.5 x 10^{—3} C) ^{2} / 2 (5
x 10^{—4} F) = 6.25 x 10^{—3}

**2**)Consider two capacitors, C1 = 6mF and C2 = 3mF in series. If the operating voltage
is 18 V, find: i) the equivalent capacitance of the combination, ii) the charge
on each plate, and iii) the p.d.s across each capacitor.

**Solution:**

i) 1/ C = 1/C1 + 1/C2 = 1/ 6mF + 1/ 3mF

Then: C **= **(C1 C2)/ (C1 + C2)** = **(6mF x 3mF)/ (9mF )

C** = **18 mF / 9mF = 2 mF

ii) Q = C V = (2 mF ) (18 V) = 36 mC

iii) Find p.d. across each capacitor:

V1 = Q/ C1 =
(36 mC )/ 6mF = 6 V

V2 = Q/ C2 =
(36 mC )/ 3 mF = 12 V

3)A parallel plate capacitor has a capacitance of 0.001 mF.

a) What p.d. is required for
a charge of 0.5 mC on each plate?

**b) **What is the total stored
energy?

**c) **If the plates are 1.0 mm
apart, find the area of each plate.

**Solutions:
**

a)
V = Q/
C = 0.5 mC / 0.001 mF = 500
V

b)
W = ½ QV* = *½
(5 x 10^{—7} C) (5 x 10^{ 2} V)* = 1.25 *x
10^{—4} * *J

c)
C** = **e_{o}** **A/d

So that: A
= Cd / e_{o } = (10^{—9} F) (10^{—3} m)/ ( 8.85 x 10^{-12} F/m)

A = 0.113 m^{ 2}

Then area of each plate: A/2 = 5.7 x 10^{ -2} m^{ 2}

4)
In a particular solar loop there are** **10^{6} weak double layers in series. Each
one is 10 cm across. If the total potential drop associated with *all
the double layers* is DV » 10 ^{9} V, estimate: i) the
potential difference across each, ii) the capacitance of each. (Assume
the same charge Q = 2 Coulomb applies to each DL)

**Solutions:
**

i)
V_{ DL}_{ } = DV /
N_{DL}_{ } = 10 ^{9} V
/ 10 ^{6} = 10 ^{3} V

ii) C _{ DL}_{ } = Q_{ DL}_{ } / V_{ DL}_{
} = 2 C / 10 ^{3} V = 2 x 10^{—3}
F

^{}

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