Physics 9702 Doubts | Help Page 175
Question 871: [Electric
field]
(a) Define electric field strength.
(b) Two flat parallel metal plates, each of length 12.0 cm, are
separated by a distance of 1.5 cm, as shown in Fig.1.
Space between the plates is a
vacuum.
The potential difference between
plates is 210 V. The electric field may be assumed to be uniform in the region
between the plates and zero outside this region.
Calculate magnitude of the electric
field strength between the plates.
(c) An electron initially travels parallel to plates along a line
mid-way between the plates, as shown in Fig.1. Speed of the electron is 5.0 ×
107 m s–1.
For the electron between the plates,
(i) determine magnitude and
direction of its acceleration,
(ii) calculate time for the electron
to travel a horizontal distance equal to the length of the plates.
(d) Use answers in (c) to determine whether the electron will hit one
of the plates or emerge from between the plates.
Reference: Past Exam Paper – June 2007 Paper 2 Q2
Solution 871:
(a) Electric field strength is defined
as the force per unit positive charge (on a small test charge)
(b) Electric field strength, E = (210 / {1.5
× 10-2} =) 1.4 ×104 N C-1
(c)
(i)
{Electric force = Eq.
Resultant force = ma. Eq = ma}
Acceleration, a = Eq / m = (1.4×104
× 1.6×10-19) / (9.1×10-31) = 2.5×1015 m s-2
(2.46×1015)
The acceleration is towards the
positive plate / upwards (and normal to the plate)
(ii)
{s = ut + ½ at2
giving s = ut + 0 since acceleration is vertical – it has no horizontal
component}
Time, t = (s
/ u = 12×10-2 / 5.0×107 =) 2.4×10-9 s
(d)
EITHER
{In this method, we are
comparing the distance the electron would travel vertically in the amount of
time calculated above.}
The vertical displacement after the
acceleration for 2.4×10-9 s is
{Initial vertical
velocity, u = 0. Using s = ut + ½ at2,}
Displacement = 0 + [½ × 2.46×1015
× (2.4×10-9)2] = 7.1 × 10-3 m (= 0.71cm)
{Since the vertical
displacement is less than half the separation [= 1.5cm / 2 = 0.75cm] of the
plates.
(Note that we are
referring to half the separation since the electron is initially incident
midway between the plates, so it only needs to move a distance of 0.75cm
vertically to reach one of the plates).}
(Since 0.71cm < 0.75cm,) The
electron will pass between the plates
OR
{In this method, we try to
compare the time the electron takes to travel halfway across (vertically) the
plates to the time it takes to travel (horizontally) a distance equal to the
length of the plate. If the electron has already travelled the horizontal
distance of the plates before moving a distance of 0.75cm vertically, then it
will not hit the plate.}
0.75 × 10-2 = ½ × 2.46 ×
1015 × t2
{Vertical distance to
travel, s = 0.75 × 10-2 m. Consider the vertical motion: s = ut + ½ at2
= 0 + ½ × 2.46 × 1015 × t2 where t is the time to travel
the 0.75m distance vertically.}
Time to travel ‘half-way across’ the
plates, t = 2.47×10-9 s
{Since the time to travel
a distance equal to the length of the plates is less than the time travel
‘half-way across’ the plates, the electron will pass between the plates before
hitting one of the plates.}
(Since 2.4ns < 2.47ns,) The
electron will pass between the plates
Question 872: [Current
of Electricity > Potentiometer]
In circuit below, P is a
potentiometer of total resistance 10 Ω and Q is a fixed resistor of resistance
10 Ω. The battery has an e.m.f. of 4.0 V and negligible internal resistance.
The voltmeter has a very high resistance.
Slider on the potentiometer is moved
from X to Y and a graph of voltmeter reading V is plotted against slider
position.
Which graph is obtained?
Reference: Past Exam Paper – June 2005 Paper 1 Q37 & November 2012
Paper 13 Q36
Solution 872:
Answer: B.
The flow of current in the circuit
is as follows. Current flows from the positive terminal of the supply to point
X – then from point X to the point where the moving slider is positioned on the
potentiometer (call this point A). From this point, the current splits: some
would flow through the voltmeter and the other would flow through resistor Q.
But since the voltmeter has a very high resistance, most (all) of the current
would flow from that point to end Y and then through resistor Q and finally
back to the negative terminal of the supply.
Thus, the voltmeter measures the p.d.
between point A and the lower junction of resistor Q.
When the slider is positioned at end
X, the total resistance connected through the voltmeter is 10Ω (due to the potentiometer) + 10Ω (due to
resistor Q) = 20Ω. In this position, the voltmeter is reading the maximum
resistance of the potentiometer + resistor Q. So, the total p.d. across them is
equal to the e.m.f. in the circuit (= 4V). [A and C
are incorrect]
As the slider is moved from X to Y,
the amount of resistance being across the voltmeter due to the potentiometer is
decreasing. At X, the resistance due to the potentiometer is 10Ω and this
decreases until it is 0Ω at Y. However, the resistor Q is always connected
across the voltmeter.
At position Y, using the potential
divider equation, the p.d. read by the voltmeter is [10 / (10+10)] × 4.0V =
2.0V. [D is incorrect]
Question 873: [Matter
> Phases]
A crystalline solid is heated at a
constant rate and change of temperature with time is shown in the graph below.
Which statement about the particles
in the material is correct?
A In the time from P to Q, the
particles are arranged randomly.
B In the time from Q to R, some
particles are arranged regularly and some particles are arranged randomly.
C In the time from R to S, the
particles are widely spaced.
D The arrangement of the particles
is the same in the time from P to S.
Reference: Past Exam Paper – June 2014 Paper 11 Q19
Solution 873:
Answer: B.
Crystalline solid is heated at a constant
rate and the change of temperature with time is shown. It can be interpreted as
follows:
Initially, it is in the crystalline
solid state (particles are orderly arranged) and remains in this state as the
temperature increases from P to Q. [A is incorrect]
The constant temperature from Q to R
indicates a change in state occurring. So, in this section, some particles are
arranged regularly and some particles are arranged randomly (both the solid is
liquids states are present). [B is correct]
From R to S, the particles are less
orderly arranged but they are not widely spaced [as they would be in the gas
state] since this is the liquid state. [C and D are
incorrect]
june 2006 q14
ReplyDeleteCheck solution 936 at
Deletehttp://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-193.html
For solution 872, the potential divider equation..Vin = (R1/R1+R2) x Vout, which one is Vin and Vout? How to know which one is Vin or Vout and which is R1 or R2?
ReplyDeleteIt is more appropriate to write it as
DeleteV1 = [R1 / (R1+R2)] x E
where V1 is the p.d. across resistor R1
E is the e.m.f.
Such a helpful website...👍👍👏👏
ReplyDeleteIn solution 872, I don't understand why all or most of the current passes through the voltmeter when it has very high resistance?
ReplyDeleteDoesn't current take the path of least resistance?
yes, the current flows from X to Y (without going through the voltmeter) as the resistance of the voltmeter has a very high resistance
DeleteAs salam u alaikum
ReplyDeleteCould you send working of 9702_w02_qp_2 Q6 in particular part (a) of that question with explanation
go to
Deletehttp://physics-ref.blogspot.com/2019/09/an-electron-travelling-horizontally-in.html