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Electrostatics
Physics: Problem of the day-18
Jun 15th
Topic: Electrostatics
An electric field line emerges from a positive point charge + q1 at an angle α to the straight line connecting it to a negative charge – q2.
Figure-1
At what angle β will the field line enter the charge – q2?
What we see in the diagram here is a cross-sectional diagram of a 3-dimensional situation. Field lines will emanate from +q1 in all directions in space. Although the presence of – q2 will prevent these field lines from being uniformly distributed, we can still assume that in the immediate proximity of +q1, the contribution to the field from –q2 will be relatively negligible, so that we can further assume that the field lines emerge from +q1 and enter –q2 in spatially homogeneous bundles. Thus, the “number of lines” emerging from +q1 and confined to the spatially conical region of vertical angle 2α will be the same as those entering –q2 and confined to the spatially conical region of vertical angle 2β.
Figure-2
The ratio number of lines emanating from +q1 in the depicted region, to the total number of lines emanating from +q1 is the ratio of the corresponding solid angles:
Therefore, the number of lines emanating from +q1 in the depicted spatial region is proportional to 1 – cos α. This immediately leads to:
Physics: Problem of the day – 12
May 5th
Topic: Electrostatics
Q: Small identical balls with equal charges are fixed at the vertices of a regular polygon with 2001 sides, each of length l. At a certain instant, one of the balls is released, and a sufficiently long time interval later, the ball adjacent to the first ball is released. The kinetic energies of the released balls are found to differ by K at a sufficiently long distance from the polygon. What is the charge q on each ball?
Physically, each ball is under the effect of potentials due to all the other balls. For example, when the first ball is released, it moves away from the polygon, since it is under the effect of repulsive forces due to all the other charges. The kinetic energy that the first ball gains when it reaches infinity, is by energy conservation the same as its electrostatic potential energy when it was in its original position.
Similarly, the kinetic energy that the second ball gains when it reaches infinity is the same as its electrostatic potential energy when it was part of the polygon, excluding the effect of the first ball.
The difference in the original potential energies of the two balls is simply q2/4πε0a, which is due to the interaction between the first and second balls. This difference is given to be equal to K. The value of q therefore becomes known.
Physics: Problem of the day – 9
May 1st
Topic: Electrostatics/Mechanics
Q. Two small balls having the same mass and charge and located on the same vertical at heights h1 and h2 are thrown in the same direction along the horizontal at the same velocity v. The first ball touches the ground at a distance l from the initial vertical.
At what height H will the second ball be at this instant?
Figure-1
The simplest way to this problem is to observe that if we treat the two balls as one system, then the Coulumb force between them becomes internal – this means that we can determine the trajectory of the system’s centre of mass just by considering the effect of gravity.
The center of mass is initially at a height (h1 + h2 )/2, and follows a trajectory given by
When x = l,
This is the y – coordinate of the center of mass – the other mass must be at double this height:
Answer
Now, its your turn to solve this problem, again, but this time without using the center of mass concept. For example, you could sit on the lower ball and observe the upper ball as it changes its vertical displacement. The acceleration of the upper ball, with respect to you, can be written as a function of the distance the upper ball is from you. You also know your total time of observation – it is (l/v) – the time it takes for you to land. Using all this information, you can set-up a differential equation and calculate the displacement of the upper ball with respect to you at any instant.
Physics: Problem of the day – 4
Apr 21st
Topic – Electrostatics/Mechanics
Q: Two small balls of mass m, bearing a charge q each, are connected by a non-conducting thread of length 2l. At a certain instant, the middle of the thread starts moving at a constant velocity v perpendicular to the direction of the thread at the initial instant.
Figure-1
Determine the minimum distance d between the balls.
This problem has a very elegant solution, if it strikes you to use the right frame of reference: that of the moving center of the thread. Note that this frame is inertial, which means there will be no pseudo-forces.
Figure-2
In this frame, at t = 0, both the two balls have a velocity of v. Also, since the two balls are charged, the initial energy stored in the system is
Now, carefully visualize the motion of the two balls from the chosen frame of reference.
Figure-3
As depicted, in this frame, the two balls have initial velocity v, which will cause the configuration of the system to change – the thread will form a “V”, with the two balls coming closer to each other. From our observation point, we’ll see that as the two balls come closer, the force of repulsion between them increases, causing the tension in the thread to increase, which in turn has an effect on the velocity of the balls, causing it to decrease.
Eventually, the increasing tension in the string will cause the velocity of the two balls (relative to our frame) to become zero – and that is the moment the balls come closest to each other. Visualize this description of the motion very carefully – that is the essence of this problem.
Figure-4
At this instant, the energy stored in the system (again, from our frame of reference, in which the balls now have zero velocity) is:
By energy conservation, we have
Elegant, isn’t it?