Tuesday, March 3, 2015

Physics 9702 Doubts | Help Page 75

  • Physics 9702 Doubts | Help Page 75



Question 402: [Current of Electricity]
Which graph best represents the way in which the current I through a thermistor depends upon the potential difference V.

Reference: Past Exam Paper – N95 / I / 14



Solution 402:
Answer: A.
The resistance of a thermistor decreases as it temperature increases.

Increasing the potential difference across a thermistor increase the current. An increase in current causes the temperature of the thermistor to increase, and hence the resistance decreases.

Ohm’s law: V = IR, giving R = V/I
1/R = I/V

If resistance R decreases, 1/R (which is equal to I/V) should increase. So, the value of I/V should increase as V increases. This is represented to graph A – joining the point results in a curve with increasing gradient.











Question 403: [Electric field]
Electric field at a certain distance from isolated alpha particle is 3.0 × 107 N C–1.
What is force on an electron when at that distance from the alpha particle?
A 4.8 × 10–12 N           B 9.6 × 10–12 N           C 3.0 × 107 N              D 6.0 × 107 N

Reference: Past Exam Paper – November 2009 Paper 11 Q29



Solution 403:
Answer: A.
The electric force on the electron is given by the product of the electric field strength, E and the charge of the electron, e.

Electric force = Ee = (3.0×107) (1.6×10-19) = 4.8×10-12 N 










Question 404: [Electric field]
Diagrams show a negative electric charge situated in uniform electric field and mass situated in uniform gravitational field.

Which row shows directions of the forces acting on the charge and on the mass?

Reference: Past Exam Paper – June 2013 Paper 13 Q12



Solution 404:
Answer: B.
In a gravitational field, the force acting on the mass is always in the direction of the field. [C and D are incorrect]

The direction of the electric field indicates the direction in which an electric force acts on a positive charge. Therefore, the direction of an electric field is from positive to negative, so that the test positive charge moves towards the negative polarity.

Hence, a negative charge would move towards a positive polarity. This is the direction opposite to the uniform electric field and is towards the positive. [A is incorrect]










Question 405: [Electric field]
Two oppositely-charged parallel plates are arranged as shown.

Electron is released from rest from surface of the negatively-charged plate.
Electron travels from negatively-charged plate towards positively-charged plate.
Which graph shows how force F on the electron varies with its distance x from the negative plate?

Reference: Past Exam Paper – June 2010 Paper 1 Q28 & Paper 12 Q26 & Paper 13 Q27



Solution 405:
Answer: D.
Electric force F = Eq

Electric field strength, E = V / d
Both the potential difference V between the plates and the separation d of the plates are constant. So, the electric field strength E is also constant.

Since the charge q of the electron is constant, the electric force F remains constant.




No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 75