• 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]