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Physics Olympiad

Maximum Electric Field of a Ring

A ring of radius (r) is located in the (x-y) plane is given a total charge (Q=2\pi R\lambda). Show that (E) is maximum when the distance (z=r/\sqrt{2}).

Discussion:

The elctric field at distance z from the centre of the ring on the axis of the ring with charge (Q=2\pi R\lambda) is given by $$ E=\frac{\lambda r}{2\epsilon_0}\frac{z}{(z^2+r^2)^{3/2}}$$

The maximum field is obtained by setting $$ \frac{dE}{dz}=0$$

This gives $$ (z^2+r^2)^{1/2}(r^2-2z^2)=0$$
Since the first factor cannot be zero for any real value of z, the second factor gives $$ z=\frac{r}{\sqrt{2}}$$

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Physics Olympiad

Tension in a Pendulum inside an Electric Field

A pendulum bob of mass (80mg) carries a charge of (2*10^{-8}C) at rest in an horizontal uniform electric field of (20,000V/m ). Find the tension in the thread of the pendulum and the angle it makes with the vertical.

Discussion:

Electric force $$ F=qE=210^{-8}20,000=410^{-4}$$
Gravitational force $$ mg= 80
10^{-6}9.8=7.8410^{-4}N$$
Balancing the horizontal and vertical components of forces $$ T \sin \theta = F$$
$$ T\cos\theta=mg$$
where (T) is the tension in the thread
Now, $$ \tan\theta=\frac{F}{mg}=\frac{410^{-4}}{7.8410^{-4}}=0.51$$
Now, (\theta=27^\circ)
To find the tension T
$$ T=\sqrt{F^2+(mg)^2}=\sqrt{(410^{-4})^2+(7.8410^{-4})^2}=8.8*10^{-4}N$$

Categories
Physics Olympiad

Total Charge on a Circular Disc

Let’s discuss a problem where we will find out the total charge on a circular disc.

The Problem:

Consider a circular disc of radius (a) whose surface density of charge at any point ((r,\theta)) is $$ \sigma(r,\theta)=\sigma_0r^2sin^2\theta$$ Find the total charge on the disc.

Solution:

Consider a surface element (rdrd\theta) on the disc about the point ((r,\theta)). The charge on is $$Q= \sigma(r,\theta)rd\theta dr$$ $$=\sigma_0\int_{0}^{a} r^3dr \int_{0}^{2\pi}sin^2\theta d\theta$$ $$=\sigma_0\frac{a^4}{4}.\pi$$$$=\pi \sigma_0\frac{a^4}{4}$$