• Physics 9702 Doubts | Help Page 69

Question 378: [Electric field]
The diagram below shows an insulating rod with equal and opposite charges at its ends, placed in a non-uniform electric field for which field lines (lines of force) are shown.

The rod experiences
A a resultant force in the plane of the paper but no couple.
B a resultant force in the plane of the paper and a couple.
C a resultant force normal to the plane of the paper but no couple.
D a resultant force normal to the plane of the paper and a couple.
E a couple but no resultant force.

Reference: Past Exam Paper – J80 / II / 18 & N86 / I / 18



Solution 378:
Answer: B.
The direction of the electric field lines shows the direction along which a free positive charge would move. So, the end of the rod with charge +Q would move in the direction of the field lines while the end of the rod with charge –Q would move in a direction opposite to the direction of the electric field.

The question states that the electric field is non-uniform. The electric force is greater where the field lines converge (where the field lines are closer). Therefore, at any instant, the electric forces acting at each end of the rod will be different in both magnitude and direction. Additionally, the 2 forces are not collinear (they do not act along the same line). Thus, the resultant force cannot be zero. [E is incorrect]

Since the electric forces act along (either in same direction or the opposite) the field lines, which are in the plane of the paper, the resultant force will also be in the plane of the paper. [C and D are incorrect]
{If the resultant force on a system is non-zero, then it is possible to represent the forces acting on the system by a single force that does not necessarily pass through the centre of mass. For example, if the net force is not zero and the net torque is perpendicular to the net force.}

The electric forces at both ends of the rod act in such a way that they cause a clockwise moment about the centre of the rod. Therefore, there is a net torque about the centre of mass of the rod. So, the 2 forces can be represented by a force acting through the centre of mass and by a couple, which is not zero. (This can be proved mathematically, but this is not in the syllabus.) [B is correct]

So, the rod would rotate while the position of the rod is itself changing.











Question 379: [Waves > Interference]
Diagram shows two loudspeakers producing sound waves that are in phase.

As student moves from X to Y, the intensity of note she hears is alternately loud and quiet.
Distance between adjacent loud and quiet regions may be reduced by
A decreasing distance d.
B increasing distance L.
C decreasing the amplitude.
D increasing the frequency.

Reference: Past Exam Paper – November 2008 Paper 1 Q28



Solution 379:

Answer: D.

The distance between adjacent loud and quiet regions is half the wavelength.

The speed of sound in air is constant. Speed of sound = fλ. Increasing the frequency f causes λ to decrease. The distance between adjacent loud and quiet regions, which is half the wavelength, also decreases. [D is correct]

Comparing with the double slit experiment:

Wavelength λ = xd / nL

where x: distance from central fringe, d: slit separation / distance between slits, n: order of fringe and L = length from slits to screen.

Since we want the distance between adjacent loud and quiet regions to decrease, x should decrease in the above formula.
x = λnL / d

Put n = 1 to obtain to make x the separation between the first set of adjacent regions.

x = λL / d
Decreasing distance d causes x to increase
Increasing distance L causes s to increase
A change in the amplitude only causes the regions to be louder or more quiet.









Question 380: [Electric field]
A point charge separated by a distance x in air from another point charge experiences a force of repulsion F. Which one of the following graphs shows how F and x are related?

Reference: Past Exam Paper – J86 / I / 17



Solution 380:
Answer: B.
The electric force F between 2 points charges is inversely proportional to the square of their separation x.

Electric force F = Q₁Q₂ / 4πϵ_ox²

The electric force F is said to be repulsive, so the 2 charges are of the same polarity. That is, the electric force has a positive sign. If the force was attractive, then the charges would have opposite polarity and thus, the electric force would have a negative sign.

Since they are INVERSELY proportional, as the separation x increases the electric force of repulsion F decreases. However, F depends (inversely) on the square of x – so, a graph of F against x or a graph of F against 1/x is NOT linear. [A and C are incorrect]

The choices given all involve straight lines. Therefore, we need to consider the general equation for a straight line and compare it with the equation obtained for the electric force.

General equation for straight line: y = mx + c
Electric force F = Q₁Q₂ / 4πϵ_ox² = [Q₁Q₂ / 4πϵ_o] (1/x²)

Comparing the 2 equations:
To obtain a straight line, F should be on the y-axis and (1/x²) should be on the x-axis. [D is incorrect]
The y-intercept c is at y = zero. [E is incorrect]
Since the electric force is positive, the graph should be above the x-axis. [B is correct]










Question 381: [Waves > Diffraction grating]
Diffraction grating with N lines per metre is used to deflect light of various wavelengths λ.
Diagram shows a relation between deflection angles θ for different values of λ in the n^th order interference pattern.

What is the gradient of graph?
A Nn                           B N / n                                    C n / N                                    D 1 / Nn

Reference: Past Exam Paper – June 2009 Paper 1 Q25



Solution 381:

Answer: A.

For diffraction grating: d sinθ_n = nλ

The diffraction grating has N lines per metre. So, the slit separation, d = 1/N.
sinθ_n / N = nλ
sinθ_n = Nnλ

The graph is shown is a straight line. Comparing with the general equation for a straight line: y = mx + c, the gradient is Nn.