## Thursday, June 24, 2021

### Solutions To Electrostatics Problems (Part 2)

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 =  Qo  e -(t/CR)

Or:   Q/  Qo   =   ½  = e -(t/CR)

2  =   e (t/CR)

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

d)      Q =   ½ Qo   = 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/ 3m

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

C = 18 mF  / 9mF  = 2 m

ii) Q = C V = (2 mF )  (18 V) = 36 m

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 =   eo A/d

So that:  A   =  Cd / eo   =  (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 106 weak  double layers in series.   Each one is 10 cm across.  If the total potential drop associated with all the double layers is D »  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 /  NDL   =  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