Download Capacitor problems in short and more Summaries Electrical Engineering in PDF only on Docsity!
CHAPTER – 31
CAPACITOR
- Given that Number of electron = 1 × 10^12 Net charge Q = 1 × 10^12 × 1.6 × 10 –19^ = 1.6 × 10–7^ C The net potential difference = 10 L.
Capacitance – C = v
q (^) = 10
1. 6 10 ^7 = 1.6 × 10 –8 F.
- A = r 2 = 25 cm 2 d = 0.1 cm
c = d
0 A =
8. 854 10 ^12 25 3. 14 = 6.95 × 10 –5 F.
- Let the radius of the disc = R Area = R^2 C = 1 D = 1 mm = 10 –3^ m
C = d
0 A
12 2 10
- 85 10 r
^
r 2 =
(^10 9) = 5998.5 m = 6 Km
- A = 25 cm^2 = 2.5 × 10 –3^ cm 2 d = 1 mm = 0.01 m V = 6V Q =?
C = d
0 A =
8. 854 10 ^12 2. 5 10 ^3
Q = CV =
8. 854 10 ^12 2. 5 10 ^3 × 6 = 1.32810 × 10–10 C
W = Q × V = 1.32810 × 10 –10^ × 6 = 8 × 10–10^ J.
- Plate area A = 25 cm 2 = 2.5 × 10–3^ m Separation d = 2 mm = 2 × 10–3^ m Potential v = 12 v
(a) We know C = d
0 A =
3
12 3 2 10
= 11.06 ×10–12 F
C =
v
q (^) 11.06 ×10 –12 (^) = 12
q
q 1 = 1.32 × 10 –10^ C. (b) Then d = decreased to 1 mm d = 1 mm = 1 × 10 –3^ m
C = d
0 A =
v
q (^) = 3
12 3 1 10
q 2 = 8.85 × 2.5 × 12 × 10 –12^ = 2.65 × 10–10^ C. The extra charge given to plate = (2.65 – 1.32) × 10 –10^ = 1.33 × 10–10^ C.
- C 1 = 2 F, C 2 = 4 F , C 3 = 6 F V = 12 V cq = C 1 + C 2 + C 3 = 2 + 4 + 6 = 12 F = 12 × 10 –6^ F q 1 = 12 × 2 = 24 C, q 2 = 12 × 4 = 48 C, q 3 = 12 × 6 = 72 C
5 cm
0.1 cm
1 mm
C 1 V C 2 C 3
The equivalent capacity. C = 2 3 1 3 1 2
1 2 3 CC CC CC
CCC
24000 = 9.23 F
(a) Let Equivalent charge at the capacitor = q
C = V
q (^) q = C × V = 9.23 × 12 = 110 C on each.
As this is a series combination, the charge on each capacitor is same as the equivalent charge which is 110 C. (b) Let the work done by the battery = W
V = q
W (^) W = Vq = 110 × 12 × 10 –6 (^) = 1.33 × 10 –3 (^) J.
8. C 1 = 8 F, C 2 = 4 F , C 3 = 4 F
Ceq = 1 2 3
2 3 1 C C C
(C C) C
8 8 = 4 F
Since B & C are parallel & are in series with A So, q 1 = 8 × 6 = 48 C q 2 = 4 × 6 = 24 C q3 = 4 × 6 = 24 C
- (a)
C 1 , C 1 are series & C 2 , C 2 are series as the V is same at p & q. So no current pass through p & q.
1 C 2
C
C
C 1 C 2
C
Cp = 2
C 1 =
4 = 2 F
And C (^) q = 2
C 2 =
6 = 3 F
C = C (^) p + C (^) q = 2 + 3 = 5 F (b) C 1 = 4 F, C 2 = 6 F, In case of p & q, q = 0
C (^) p = 2
C 1 =
4 = 2 F
Cq = 2
C 2 =
6 = 3 F
& C = 2 + 3 = 5 F
C & C = 5 F
The equation of capacitor C = C + C = 5 + 5 = 10 F
12 V
A^8 F 4 F B^ C 4 F
A (^) B
C 1
C 2
C 1
C 2
C 1 = 4 C 2 = 6
A
p
B C 1 C 1
C 1 C 1
q
R
S
C 2
C 2
C 2
C (^2)
30 F
V = 12 V
20 F 40 F
(a) Capacitor = 4 8
and 6 3
= 2 F
(i) The charge on the capacitance 3
8 F
Q =
8 × 50 =
The potential at 4 F = 3 4
at 8 F = 3 8
The Potential difference = 6
50 V
(ii) Hence the effective charge at 2 F = 50 × 2 = 100 F
Potential at 3 F = 3
(^100) ; Potential at 6 F = 6
Difference = 6
50 V
The potential at C & D is 3
50 V
(b) S
R
q
P =
(^1) = It is balanced. So from it is cleared that the wheat star bridge balanced. So
the potential at the point C & D are same. So no current flow through the point C & D. So if we connect another capacitor at the point C & D the charge on the capacitor is zero.
- Ceq between a & b
= 1 2
3 1 2 1 2
1 2 C C
C CC
C C
CC
1 2
3 1 2 C C
C 2 CC
(The three are parallel)
- In the figure the three capacitors are arranged in parallel.
All have same surface area = a = 3
A
First capacitance C 1 = 3 d
0 A
2 nd^ capacitance C 2 = 3 (b d)
0 A
3 rd^ capacitance C 3 = 3 ( 2 b d)
0 A
Ceq = C 1 + C 2 +C (^3)
8 F
B
C A
4 F
3 F D 6 F 50
A
C
B
D
6 F
8 F
3 F
4 F (^8) F C
D
50
4 F
3 F 6 F
C 1
C 2 a C (^3) b
C 2
C 1
C 1 C 2 /C 1 +C 2
a (^) C 3 b
C 1 C 2 /C 1 +C 2
b
B C D
a
A
d
a
a
b
E
3 d
0 A+
3 (b d)
0 A
3 ( 2 b d)
0 A
2 b d
b d
d
0 A
d(b d)( 2 b d )
(b d)( 2 b d) ( 2 b d)d (b d)d 3
0 A
= ^
3 d(b d)( 2 b d )
0 A^3 d^26 bd^2 b^2
- (a) C = In(R /R)
2 L
2 1
In 2
e 3. 14 8. 85 10 ^2 10 ^1 [In2 = 0.6932]
= 80.17 × 10–13^ 8 PF
(b) Same as R 2 /R 1 will be same.
- Given that C = 100 PF = 100 × 10 –12^ F Ccq = 20 PF = 20 × 10–12^ F V = 24 V q = 24 × 100 × 10–12^ = 24 × 10– q 2 =? Let q 1 = The new charge 100 PF V 1 = The Voltage. Let the new potential is V 1 After the flow of charge, potential is same in the two capacitor
V 1 = 2
2 C
q (^) = 1
1 C
q
2
1 C
q q = 1
1 C
q
10 24 10
24 10 q
12
1 100 10
q ^
= 24 × 10 –10^ – q (^) 1 = 5
q 1
= 6q 1 = 120 × 10–
= q 1 = 6
120 ×10–10 = 20 × 10–
V1 =
1
1 C
q (^) = 12
10 100 10
= 20 V
Initially when ‘s’ is not connected,
Ceff = q 3
2 C = 50
2 C = 104
5 = 1.66 × 10–4 C
After the switch is made on, Then C (^) eff = 2C = 10– Q = 10 –5^ × 50 = 5 × 10 – Now, the initial charge will remain stored in the stored in the short capacitor Hence net charge flowing = 5 × 10 –4^ – 1.66 × 10 –4^ = 3.3 × 10 –4^ C.
A
S (^) /
Hence E = (^2) 0 a
q
1 2 0
0 2 (d d) a
a V
(d d)
V
Substituting the data in the known equation, we get, 2
d (^1) = 2
2
1 2 u
a (d d)m
e V 2
u^2 = dm(d d)
Vea 1 1 2
2
u =
1 / 2
1 1 2
2 dm(d d)
Vea
- The acceleration of electron ae = Me
qeme
The acceleration of proton = Mp
qpe (^) = ap
The distance travelled by proton X = 2
(^1) apt (^2) …(1)
The distance travelled by electron …(2)
From (1) and (2) 2 – X = 2
(^1) a c t^2 x =^2
(^1) a c t^2
c
p a
a 2 x
x (^)
c
c
p
p
M
qF
M
qE
p
c M
M
2 x
x (^)
31
- 67 10
9. 1 = 5.449 × 10 –
x = 10.898 × 10 –4^ – 5.449 × 10–4^ x
x =
- 0005449
10. 898 10 ^4 = 0.
- (a)
As the bridge in balanced there is no current through the 5 F capacitor So, it reduces to similar in the case of (b) & (c) as ‘b’ can also be written as.
Ceq = 2 6
6 12 = 2.25 F
- (a) By loop method application in the closed circuit ABCabDA
–12 + 4 F
Q
2 F
Q
2 F
2 Q 11
In the close circuit ABCDA
–12 + 4 F
Q Q
2 F
Q 1
From (1) and (2) 2Q + 3Q 1 = 48 …(3) And 3Q – q 1 = 48 and subtracting Q = 4Q 1 , and substitution in equation
2 cm qe epx
qp E
e–x E E
6 F
A (^) B
1 F 3 F
2 F
5 F A^ B
1
2
3
6 1 F
6 F
3 F
2 F
5 F
B 2 F^ C^2 F
4 F
b
4 F
Q
a
A D
Q
(Q – Q 1 )
2Q + 3Q 1 = 48 8 Q 1 + 3Q 1 = 48 11Q 1 = 48, q 1 = 11
Vab = 4 F
Q 1
12 V
(b)
The potential = 24 – 12 = 12
Potential difference V = 2 4
48 = 8 V
The Va – Vb = – 8 V (c)
From the figure it is cleared that the left and right branch are symmetry and reversed, so the current go towards BE from BAFEB same as the current from EDCBE.
The net charge Q = 0 V = C
Q =
C
(^0) = 0 Vab = 0
The potential at K is zero. (d)
The net potential = Netcapacitance
Net charge = 7
72 10.3 V
Va – Vb = – 10.3 V
- (a)
By star Delta conversion
Ceff =
^
^
11 F
12 V
2 F 4 F
24 V
a
b
2 F
12 V 4 F
24 V
a
b
Right A C
D
2 F
2 V a
2 F b
Left B 2 V
F E
a b 24 V
4 F 2 F
1 F
6 V
12 V
3/
4/8 12/
3 F 1 F 3 F 1 F
3/
1/2 3/
3 F
4/
1 F
4 F 12/
1 F
3/
3 F
= C 5 and C 1 are in series
Ceq = 2 2
This is parallel to C 6 = 1 + 1 = 2
Which is series to C 2 = 2 2
Which is parallel to C 7 = 1 + 1 = 2
Which is series to C 3 = 2 2
Which is parallel to C 8 = 1 + 1 = 2
This is series to C 4 = 2 2
Let the equivalent capacitance be C. Since it is an infinite series. So, there will be negligible change if the arrangement is done an in Fig –
Ceq = 2 C
2 C
+ 1 C =
2 C
2 C 2 C
(2 + C) × C = 3C + 2
C^2 – C – 2 = 0
(C –2) (C + 1) = 0
C = –1 (Impossible) So, C = 2 F
= C and 4 f are in series
So, C 1 = 4 C
4 C
Then C 1 and 2 f are parallel C = C 1 + 2 f
2 4 C
4 C
4 C
4 C 8 2 C
= C
4C + 8 + 2C = 4C + C 2 = C 2 – 2C – 8 = 0
C =
C =
2 (^6) = 4 f
The value of C is 4 f
1
2 f 2 f^2 f^2 f 2 3 4
5 6 ^7 ^8
A
B
Fig -
A
B
C
A
B
Fig – 1 F
2 f
A
B
C 2 f (^2) f 2 f
4 f^4 f^4 f
B
A c (^) C
4 f
2 f
- q 1 = +2.0 × 10 –8^ c q 2 = –1.0 × 10 –8^ c C = 1.2 × 10 –3^ F = 1.2 × 10 –9^ F
net q = 2
q 1 q (^2) = 2
3. 0 10 ^8
V =
c
q (^) = 9
8
- 2 10
= 12.5 V
- Given that Capacitance = 10 F Charge = 20 c
The effective charge = 2
20 0 = 10 F
C =
V
q (^) V = C
q (^) = 10
10 = 1 V
- q 1 = 1 C = 1 × 10 –6^ C C = 0.1 F = 1 × 10–7^ F q 2 = 2 C = 2 × 10 –6^ C
net q = 2
q 1 q (^2) = 2
( 1 2 ) 10 ^6 = – 0.5 × 10–6 C
Potential ‘V’ = c
q (^) = 7
7 5 10
= – 5 V
But potential can never be (–)ve. So, V = 5 V
- Here three capacitors are formed And each of
A = 12 0
f.m.
d = 4 mm = 4 × 10–3^ m Capacitance of a capacitor
C =
d
0 A =
3
0
12 0 4 10
= 24 × 10 –9^ F.
As three capacitor are arranged is series
So, Ceq = q
C =
24 10 ^9 = 8 × 10–
The total charge to a capacitor = 8 × 10 –9^ × 10 = 8 × 10 –8^ c The charge of a single Plate = 2 × 8 × 10 –8^ = 16 × 10 –8^ = 0.16 × 10 –6^ = 0.16 c.
- (a) When charge of 1 c is introduced to the B plate, we also get 0.5 c charge on the upper surface of Plate ‘A’. (b) Given C = 50 F = 50 × 10–9^ F = 5 × 10 –8^ F Now charge = 0.5 × 10 –6^ C
V = C
q (^) = 5 10 F
5 10 C
8
7
= 10 V
- Here given, Capacitance of each capacitor, C = 50 f = 0.05 f Charge Q = 1 F which is given to upper plate = 0.5 c charge appear on outer and inner side of upper plate and 0.5 c of charge also see on the middle. (a) Charge of each plate = 0.5 c Capacitance = 0.5 f
x x 10^10
+20c
- - - - - -
x 10
0.5 C
B
1 C 0.5^ C 0.5 C
0.5 C
0.5 C
0.5 C
0.5 C
- Initial charge stored = C × V = 12 × 2 × 10 –6^ = 24 × 10 –6^ c Let the charges on 2 & 4 capacitors be q 1 & q 2 respectively
There, V = 1
1 C
q
2
2 C
q 2
q (^1) = 4
q (^2) q 2 = 2q 1.
or q 1 + q 2 = 24 × 10 –6^ C q 1 = 8 × 10 –6^ c q 2 = 2q 1 = 2 × 8 × 10 –6^ = 16 × 10 –6^ c
E 1 = (1/2) × C 1 × V 12 = (1/2) × 2 ×
2 2
= 16 J
E 2 = (1/2) × C 2 × V 22 = (1/2) × 4 ×
2 4
= 8 J
- Charge = Q Radius of sphere = R Capacitance of the sphere = C = 4 0 R
Energy = C
Q
4 R
Q
0
2
8 R
Q
0
2
42. Q = CV = 4 0 R × V
E =
C
q 2
(^1 2) [ ‘C’ in a spherical shell = 4 0 R]
E =
4 2 R
16 R V
0
(^222) 0
2
0 RV^2 [‘C’ of bigger shell = 4^ 0 R]
- = 1 × 10 –4^ c/m 2 a = 1 cm = 1 × 10–2^ m a^3 = 10 –6^ m
The energy stored in the plane = 0
2 2
12
42
- 85 10
The necessary electro static energy stored in a cubical volume of edge 1 cm infront of the plane
= 3 0
2 a 2
= 265 × 10 –6 = 5.65 × 10 –4 J
- area = a = 20 cm^2 = 2 × 10 –2^ m 2 d = separation = 1 mm = 10 –3^ m
Ci = (^3)
3 0 10
= 2 0 C = 3
3 0 2 10
0
qi = 24 0 q = 12 0 So, q flown out 12 0. ie, q (^) i – qf. (a) So, q = 12 × 8.85 × 10 –12^ = 106.2 × 10 –12^ C = 1.06 × 10 –10^ C (b) Energy absorbed by battery during the process = q × v = 1.06 × 10–10^ C × 12 = 12.7 × 10–10^ J (c) Before the process Ei = (1/2) × Ci × v^2 = (1/2) × 2 × 8.85 × 10 –12^ × 144 = 12.7 × 10–10^ J After the force Ei = (1/2) × Cf × v 2 = (1/2) × 8.85 × 10 –12^ × 144 = 6.35 × 10 –10^ J (d) Workdone = Force × Distance
= A
q 2
0
2
= 1 × 10^3 = 3
0
0 0 3 2 10
(e) From (c) and (d) we have calculated, the energy loss by the separation of plates is equal to the work done by the man on plate. Hence no heat is produced in transformer.
R
Q
- (a) Before reconnection C = 100 f V = 24 V q = CV = 2400 c (Before reconnection) After connection When C = 100 f V = 12 V q = CV = 1200 c (After connection) (b) C = 100, V = 12 V q = CV = 1200 v
(c) We know V = q
W
W = vq = 12 × 1200 = 14400 J = 14.4 mJ The work done on the battery. (d) Initial electrostatic field energy Ui = (1/2) CV 12 Final Electrostatic field energy U = (1/2) CV 22 Decrease in Electrostatic Field energy = (1/2) CV 12 – (1/2) CV 22 = (1/2) C(V 12 – V 22 ) = (1/2) × 100(576 –144) = 21600J Energy = 21600 j = 21.6 mJ (e)After reconnection C = 100 c, V = 12 v The energy appeared = (1/2) CV 2 = (1/2) × 100 × 144 = 7200 J = 7.2 mJ This amount of energy is developed as heat when the charge flow through the capacitor.
- (a) Since the switch was open for a long time, hence the charge flown must be due to the both, when the switch is closed. Cef = C/
So q = 2
E C
(b) Workdone = q × v = E 2
EC =
E 2 C
(c) E (^) i = E^2 2
C
E 2 C
Ef = (1/2) × C × E 2 = 2
E 2 C
Ei – Ef = 4
E 2 C
(d) The net charge in the energy is wasted as heat.
- C 1 = 5 f V 1 = 24 V q 1 = C 1 V 1 = 5 × 24 = 120 c and C 2 = 6 f V 2 = R q 2 = C 2 V 2 = 6 × 12 = 72 Energy stored on first capacitor
Ei = 1
2 1 C
q 2
(^1) 2 = 1440 J = 1.44 mJ
Energy stored on 2 nd^ capacitor
E 2 = 2
2 2 C
q 2
(^1) 2 = 432 J = 4.32 mJ
E
C C
- Initial charge stored = 50 c Let the dielectric constant of the material induced be ‘k’. Now, when the extra charge flown through battery is 100. So, net charge stored in capacitor = 150 c
Now C 1 = d
0 A or d
A
V
q (^1) (^0) …(1)
C 2 =
d
0 Ak or, V
q (^2) d
0 Ak …(2)
Deviding (1) and (2) we get k
q
q 2
k
(^50) k = 3
- C = 5 f V = 6 V d = 2 mm = 2 × 10–3^ m. (a) the charge on the +ve plate q = CV = 5 f × 6 V = 30 c
(b) E = d
V =
2 10 m
6 V
^3 = 3 × 10^
3 V/M
(c) d = 2 × 10 –3^ m t = 1 × 10 –3^ m
k = 5 or C = d
0 A 5 × 10 –6 = 9
3
12 10 2 10
8. 85 A 10
A =
When the dielectric placed on it
C 1 =
k
d t t
0 A
3 3
12 4
^
12 4 6 10
5 = 0.00000833 = 8.33 F.
(d) C = 5 × 10 –6^ f. V = 6 V Q = CV = 3 × 10 –5^ f = 30 f C = 8.3 × 10 –6^ f V = 6 V Q = CV = 8.3 × 10–6^ × 6 ≈ 50 F charge flown = Q – Q = 20 F
- Let the capacitances be C 1 & C 2 net capacitance ‘C’ = 1 2
1 2 C C
CC
Now C 1 = 1
0 1 d
Ak C 2 = 2
0 2 d
Ak
C =
2
0 2 1
0 1
2
0 2 1
0 1
d
Ak d
Ak
d
Ak d
Ak
12
0 12 21
12
0 12
dd
kd kd A
dd
A kk = (^33)
12 2 6 4 10 4 6 10
= 4.425 × 10–11^ C = 44.25 pc.
- A = 400 cm 2 = 4 × 10–2^ m 2 d = 1 cm = 1× 10–3^ m V = 160 V t = 0.5 = 5 × 10–4^ m k = 5
50 c
k=4 (^) 4 mm
6 mm
C 2
C 1
1 cm^2
C =
k
d t t
0 A
3 4 4
12 2
^
3
4
- (a) Area = A Separation = d
C 1 = d/ 2
0 Ak (^1) C 2 =^ d/ 2
0 Ak (^2)
C =
1 2
1 2 C C
CC
d
2 Ak d
2 Ak
d
2 Ak d
2 Ak
0 1 0 2
0 1 0 2
2 0 1 2
2
12
2 0
d
( 2 A)kd kd
d
( 2 A)kk
^
d(k k )
2 kk A 1 2
120
(b) similarly
1 2 C 3
C
C
C
d
3 Ak
d
3 Ak
d
3 Ak
0 1 2 k 3
k
k
3 A
d (^) =
23 13 12 0 kkk
kk kk kk 3 A
d
C =
d(kk kk kk)
3 Akkk 12 2 3 13
0 12 3
(c) C = C 1 + C (^2)
d
k 2
A
d
k 2
A
= (k k ) 2 d
A
Consider an elemental capacitor of with dx our at a distance ‘x’ from one end. It is constituted of two capacitor elements of dielectric constants k 1 and k 2 with plate separation xtan and d –xtan respectively in series
1 dc 2
dc
dcR
k(bdx)
d xtan k(bdx)
xtan 0 2 01
dcR =
2 1
0
k
(d xtan ) k
xtan
bdx
or C (^) R = 0 bk 1 k (^2) kd (k k )xtan
dx 2 1 2
= tan (k k )
bkk 1 2
0 12
(^) [log ek^2 d+ (k^1 –k 2 ) x tan]a
tan (k k )
bkk 1 2
0 12
(^) [log ek^2 d+ (k^1 –k 2 ) a tan^ – loge k^2 d]
tan = a
d (^) and A = a × a
K (^1)
K 2
B
A
X
dx
d d – x tan k 2 x tan
k 1 d dc
dc^1
- Capacitance = 100 F = 10 –4^ F P.d = 30 V (a) q = CV = 10–4^ × 50 = 5 × 10 –3^ c = 5 mc Dielectric constant = 2. (b) New C = C = 2.5 × C = 2.5 × 10 –4^ F
New p.d = (^1) c
q (^) [’q’ remains same after disconnection of battery]
3
- 5 10
= 20 V.
(c) In the absence of the dielectric slab, the charge that must have produced C × V = 10 –4^ × 20 = 2 × 10–3^ c = 2 mc (d) Charge induced at a surface of the dielectric slab = q (1 –1/k) (where k = dielectric constant, q = charge of plate)
= 5 × 10–3^
1 1 = 5 × 10 –3^ ×
(^3) = 3 × 10 –3 (^) = 3 mc.
- Here we should consider a capacitor cac and cabc in series
Cac = k(c a)
4 0 ack
Cbc = (b c)
4 0 bc
Cbc
Cac
C
4 bc
(b c) 4 ack
(c a) 0 0
k 4 abc
b(c a) ka(b c) 0
C =
ka(b c) b(c a)
4 0 kabc
- These three metallic hollow spheres form two spherical capacitors, which are connected in series. Solving them individually, for (1) and (2)
C 1 = b a
4 0 ab
(^) ( for a spherical capacitor formed by two spheres of radii R 2 > R^1 )
C = 2 2
0 2 1 R R
4 RR
Similarly for (2) and (3)
C 2 = c b
4 0 bc
Ceff = 1 2
1 2 C C
CC
(b a)(c b )
4 ab(c b) bc(b a)
(b a)(c a)
( 4 )abc
0
2 2 0
abc ab bc abc
4 abc 2 2
0 2
b(c a)
4 abc 2
0 2
c a
4 0 ac
- Here we should consider two spherical capacitor of capacitance cab and cbc in series
Cab = (b a)
4 0 abk
Cbc = (c b)
4 0 bc
C
b
C
b
a
C a
b
C
Cbc
Cab
C
4 bc
(c b) 4 abk
(b a) 0 0
k 4 abc
c(b a) ka(c b) 0
C =
c(b a) ka(c b)
4 0 kabc
- Q = 12 c V = 1200 V
d
v (^) = 3 × 10– m
v
d = (v/d)
V =
= 4 × 10^
–4 (^) m
c = v
Q =
12 10 ^6 = 10 –8 (^) f
C =
d
0 A = 10 –8 (^) f
A =
0
108 d
4
8 4
- 854 10
0.45 m 2
- A = 100 cm 2 = 10 –2^ m 2 d = 1 cm = 10 –2^ m V = 24 V (^0)
The capacitance C = d
0 A =
2
12 2 10
^
= 8.85 × 10–
The energy stored C 1 = (1/2) CV 2 = (1/2) × 10 –12^ × (24) 2 = 2548.8 × 10–
The forced attraction between the plates = d
C 1 =
2
12 10
= 2.54 × 10–7^ N.
We knows
In this particular case the electricfield attracts the dielectric into the capacitor with a force 2 d
0 bV 2 (k 1 )
Where b – Width of plates k – Dielectric constant d – Separation between plates V = E = Potential difference. Hence in this case the surfaces are frictionless, this force is counteracted by the weight.
So, 2 d
0 bE 2 (k 1 )= Mg
M =
2 dg
0 bE 2 (k 1 )
M
d K
