Download If V0 is the supply voltage d is the separation between plates P and Q

CHAN SHU KUI MEMORIAL SCHOOL
S.7 PHYSICS TEST 2
Name:
1. A capacitor C1 is given an initial charge Q. It is then shares its charge with an uncharged
capacitor C2 as shown in Figure 1.
+Q1
C1
+Q2
C2
Figure 1.
-Q1
-Q2
(a)
Find the final charges Q1 and Q2 on the capacitors in terms of C1 , C2 and Q.
(4 marks)
(b)
Hence, find the final charge Q2 on C2 if the capacitance of C2 is much greater than that of
C1 .
(2 marks)
2. A capacitor is formed by two parallel conducting plates P and Q each of area 0.2m 2. They
are 15 mm apart and are connected to a 120 V supply as shown in Figure 2.
15 mm
Figure 2.
(a)
P
Q
120 V
Determine
( i ) the electric field strength E between the plates,
(3 marks)
( ii) the surface charge density on each plate,
(2 marks)
(b)
(iii) the total charge on each plate and,
(2 marks)
(iv) the energy stored in the electric field between the plates.
(2 marks)
Suppose another two parallel conducting plates A and B are inserted between plates
P and Q as shown in Figure 3, so that the distance between adjacent plates is 5 mm.
5 mm
Figure 3.
P
5 mm
A
5 mm
B
Q
120 V
The potential difference between plates P and Q is still maintained at 120 V and the
thicknesses of plates A and B can be neglected.
(i)
What is the new capacitance of the capacitor?
(2 marks)
(ii)
Hence, or otherwise, determine the amount of charges and the electric field
strength in the capacitor.
(2 marks)
Solutions to Test 2
1.
(a)
Consenation of Charge: Q = Q1 + Q2 ---------(1)
Q
Q
Same p.d. across Co and C: 1  2 --------(2)
C1 C 2
(1) and (2)  Q1 
C1
C2
Q and Q2 
C1  C 2
C1 C 2
(b) C2>>C1  Q2 = Q
1M
1M
2M
1M+1M
2. (a)(i) If V0 is the supply voltage d is the separation between plates P and Q
E= V0/d = 120/0.015
1M
-1
=8000 Vm
1A
(i) The amount of the charges on plates P and Q are equal in magnitude but opposite in sign. The
charges on plate P is positive whereas those on plate Q is negative. If σ and Q0 are the
magnitude of the surface charge density and the total charge on each plates respectively.
1M
1A
1M
1A
1M
1A
(b ) (i) When two parallel conducting plates A and B are inserted into the gap the set-up is equivalent
to three capacitors connected in series , The capacitance of each is given by
CpA = CAB =CBQ = (εoA)/(d/3) =3 C0
1M
Where C0 is the capacitance before plates A and B are inserted. The equivalent capacitance becomes
CPQ =(3C0) / 3 = C0 =Q0/V0 =1.416 × 10-8 =1.18 V0 10-10 F
1A+1A
(ii) Sine the potential difference between plates P and Q remains 120 V.
Q = CPQ V0
= ( 1.18 × 10-10)(120)
=1.416 × 10-8 C
The potential difference between the adjacent plates is now given by
VpA = VAB = VBQ
1M
= 1/3VpQ = 1/3V0 (since V∝1/C)
By the equation E = V/d we have
EpA = EAB = EBQ
= (1/3V0)/(1/3d)
= V0/d = 8000Vm-1
Therefore there is no change of the electric field strength in the capacitor.
1M
1A