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Physics Olympiad

Multipole Expansion of a Potential

Q. An insulating rod running from (z=-a to z=a) carries indicated line charges. Determine the leading term in the multipole expansion of the potential is given by $$ \lambda=kcos(\frac{\pi z}{2a})$$ where k is a constant.

Solution:
Potential due to multipole expansion is given by $$ V(\vec{r})=\frac{1}{4\pi\epsilon_0}\sum_{0}^{\inf} \frac{P_n(cos\theta)}{r^{n+1}}I_n$$ where (I_n=\int_{-a}^{+a}z^n\lambda(z)dz)
Inserting the value of charge density in the integral to obtain: $$I_0=k\int_{-a}^{a} cos(\frac{\pi z}{2a})dz=\frac{2ak}{\pi}sin(\frac{\pi z}{2a}$$
Applying limits (z=-a to z=a) $$ I_0= \frac{2ak}{\pi}[sin (\pi/2)+sin(\pi/2)]=\frac{4ak}{\pi}$$
The potential $$ V(r,\theta)=\frac{1}{4\pi\epsilon_0}(4ak/pi)$$
.

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Physics Olympiad

Removed Charge Problem (Kvpy ’10)

Let’s discuss a problem based on Removed Charge from Kishore Vaigyanik Protsahan Yojana, KVPY, 2010. First, try the problem yourself, then read the solution here.

The Problem:

(12) positive charges of magnitude (q) are placed on a circle of radius (R) in a manner that they are equally spaced. A charge (Q) is placed at the center. If one of the charges (q) is removed, then the force on (Q) is
(A) zero
(B)$$ \frac{qQ}{4\pi\epsilon_0R^2}$$ away from the position of the removed charge

(C) $$ \frac{11qQ}{4\pi\epsilon_0R^2}$$ away from the position of the removed charge
(D)
$$ \frac{qQ}{4\pi\epsilon_0R^2}$$ towards from the position of the removed charge

Discussion:

If one of the charges is removed, then the net force on Q is $$ \frac{qQ}{4\pi\epsilon_0R^2}$$ towards the position of removed charge

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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

Electric Field of a Charged Hemisphere

A total charge (q) is spread uniformly over the inner surface of a non-conducting hemispherical cup of inner radius (a). Calculate the electric field.

Discussion:

Consider a circular strip symmetric about (z-axis) of radius (r) and width (ad\theta)
The charge on the strip is $$ dq=q\frac{2\pi r ad\theta}{2\pi a^2}=\frac{qr d\theta}{a}=q sin\theta d\theta$$
(a) At the centre of the hemisphere, the (x-component) of the field will be cancelled for reasons of symmetry. The entire field will be contributed by the (z-component) alone.
$$ dE=dE_z= \frac{q sin\theta d\theta cos\theta}{4\pi\epsilon_0 a^2}
$$
Therefore, $$ E=\int dE_z=\frac{q}{4\pi\epsilon_0 a^2}\int_{0}^{\pi/2}sin \theta cos\theta d\theta =\frac{q}{8\pi \epsilon_0 a^2}$$

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Physics Olympiad

Electric Field from Electric Potential

If the electric potential is given by (\chi=cxy), calculate the electric field.

Discussion:

$$ E_x=-\frac{\partial\chi}{\partial x}=-cy$$
$$ E_y=\frac{\partial \chi}{\partial y}=-cx$$
Hence electric field $$ \vec{E}=-c(y\hat{i}+x\hat{j})$$

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Physics Olympiad

Breakdown Voltage of a Sphere

 

What is the maximum charge that can be given to a sphere of diameter (10cm) if the breakdown voltage of air is (2\times10^4 V/cm)?

Discussion:

Dielectric breakdown is when current flows through an electrical insulator when the voltage applied across it exceeds the breakdown voltage. This results in the insulator becoming electrically conductive.

The breakdown voltage of air is (2\times10^4 V/cm)

For a sphere, $$ V= \frac{q}{4\pi \epsilon_0 r}$$
We can find charge $$ q= 4\pi \epsilon_0 r V$$ $$ = \frac{(5)(2\times10^4)}{9\times10^9}$$ $$=1.11\times 10^{-5}$$

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Physics Olympiad

Infinite Number of Charges Problem

Let’s discuss a problem on Infinite Number of Charges from Physics Olympiad. Try the problem first, and then read the solution here.

The Problem: Infinite Number of Charges

An infinite number of charges, each equal to (q), are placed along the x-axis at (x=1),(x=2),(x=4),(x=8) etc. Find the potential and electric field at the point x=0 due to the set of the charges.

Discussion:

An infinite number of charges, each equal to (q), are placed along the x-axis at (x=1),(x=2),(x=4),(x=8) etc.

Electric potential $$ V=\frac{1}{4\pi\epsilon_0}(\frac{q}{1}+\frac{q}{2}+\frac{q}{4}+\frac{q}{8}+…)$$ $$=\frac{1}{4\pi\epsilon_0}(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+….)$$
The terms in brackets form a geometric progression of infinite terms whose sum is $$ S= \frac{a}{1-r}=\frac{1}{1-1/2}=2$$

Hence the potential $$ V=q/2\pi\epsilon_0$$

Some useful links:

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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
$$ T=\sqrt{F^2+(mg)^2}=\sqrt{(410^{-4})^2+(7.8410^{-4})^2}=8.8*10^{-4}N$$

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Physics Olympiad

Three Charged Spheres

Let’s discuss a problem where we will find the charge of the sphere with the help of the three charged spheres problem. Try before reading the solution.

The Problem:

Consider three concentric metallic spheres \(A\), \(B\) and \(C\) of radii of \(a\), \(b\), \(c\) respectively where a<b<c . \(A\) and \(B\) are connected whereas C is grounded The potential of the middle sphere \(B\) is raised to \(V\) then what is the charge on the sphere \(C\)?

Solution:

Three concentric metallic spheres \(A\), \(B\) and \(C\) have radii of \(a\), \(b\), \(c\) respectively where a<b<c . \(A\) and \(B\) are connected whereas C is grounded The potential of the middle sphere \(B\) is raised to \(V\).

$$ V=\frac{Kq}{b}+\frac{KQ}{c}$$ $$
\Rightarrow \frac{k(q+Q)}{c}=0$$
$$\Rightarrow q+Q=0$$
$$\Rightarrow q=-Q
$$
$$ \frac{k(-Q)}{b}+\frac{KQ}{c}=V$$
$$\Rightarrow KQ(\frac{1}{c}-\frac{1}{b})=V$$
$$ Q=\frac{bcV}{(b-c)}4\pi\epsilon_0$$

Useful links:

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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 $$Q= \sigma(r,\theta)rd\theta dr$$ $$=\sigma_0\int_{0}^{a} r^3dr \int_{0}^{2\pi}sin^2\theta d\theta$$ $$=\sigma_0\frac{a^4}{4}.\pi$$$$=\pi \sigma_0\frac{a^4}{4}$$