Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
1) See attached figure
The relationship between charge and current is:
where
i is the current
Q is the charge
t is the time
Therefore, the current is the rate of change of the charge passing through a given point over time.
This means that for a graph of charge over time, the current is just equal to the slope of the graph.
For the graph in this problem:
- Between t = 0 and t = 2 s, the slope is

therefore the current is
i = 25 A
- Between t = 2 s and t = 6 s, the slope is

therefore the current is
i = -25 A
- Between t = 6 s and t = 8 s, the slope is

therefore the current is
i = 25 A
The figure attached show these values plotted on a graph.
2)
The previous equation can be rewritten as
This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.
Here we have the current vs time graph, so we gave to find the area under it.
The area of the first triangle is:

While the area of the second square is

So, the total area (and the total charge) is

Most appropriate unit of measurement should be centimetres, cm.
Answer:
The factor of the diameter is 0.95.
Explanation:
Given that,
Power of old light bulb = 54.3 W
Power = 60 W
We know that,
The resistance is inversely proportional to the diameter.

The power is inversely proportional to the resistance.


We need to calculate the factor of the diameter of the filament reduced
Using relation of power and diameter

Put the value into the formula



Hence, The factor of the diameter is 0.95.