(EM Griffiths) Problem 2.2 (b)

Question
Report part(a), only this time make the right hand charge -q instead of +q. 

Answer

As we can easily conclude from the diagram, vertical components here cancel each other, only surviving components are horizontals. Consider this two dimensional problem is confined in the z-x plane.

So the horizontal resultant force field along x-axis is, 
$E_x = 2q sin\theta/4\pi\epsilon_0 r^2$

From the diagram,
$sin\theta = d/2r $
and $r = \sqrt{z^2 + (d/2)^2}$

Therefore net force field is as follows, 
 $E = qd\hat{x}/4\pi\epsilon_0 (z^2 + (d/2)^2)^{3/2}$

For z>> d the field goes like, 
$E = qd\hat{x}/4\pi\epsilon_0 z^3$
This is as we know is the field of a dipole. 

(EM Griffiths) Problem 2.2 (a)

Question
Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges, q, a distance d apart (as shown in the figure). Check that your result is consistent with what you'd expect when z>>d.

Answer

As it is given in the problem, q charges are of same polarity. Naturally the directions of the force fields are as shown in figure, along AE and BD. It is clear from the figure that the horizontal components (hc) of the force fields cancel each other out, only surviving components are verticals. They are directed along the same direction. So the net force field is the  simple algebraic sum of the two vertical components.

Therefore the resultant force field is as follows,

$E_z = {2q cos \theta}/{4\pi \epsilon_0 r^2}$

The factor 2 comes from the fact that, vertical forces are also equal in magnitude.

Here, 

$r = \sqrt {z^2+(d/2)^2}$ 

and

$ cos\theta = sin (90-\theta) = z/r $

As we can see in the given figure.

So, 

$E= 2qz\hat{z}/4\pi\epsilon_0 (z^2+(d/2)^2)^{3/2} $

When z>>d, field point is so far away that from it the source charges almost look like a point charge of magnitude 2q. 

Now the net field is, 

$E= 2qz\hat{z}/4\pi\epsilon_0 z^3 $

$E= 2q\hat{z}/4\pi\epsilon_0 z^2 $

We just set, 

$ d \xrightarrow{} 0 $

(EM Griffiths) Problem 2.1 (d)

Question
If one of the 13 q's is removed, What is the force on Q?

Answer
Similar to the problem 2.1(b), the answer is

$F_Q = Qq/4\pi\epsilon_or^2$

Where r is the distance between charge Q and q.

Reasoning is easy. One missing charge breaks the symmetry of the problem. As a result of that charges sitting at corners other than the vacant site impose resultant coulomb force of amount $F_Q$ upon the test charge Q. We can easily prove it analytically (by using calculus). 

(EM Griffiths) Problem 2.1 (c)

Question

Now 13 equal charges, q, are placed at the corners of a regular 13-sided polygon. What is the force on a test charge Q at the center?

Answer
This is a little complicated scenario. There are no pairs of charges sitting at opposite sites. It's a bit delicate to describe it with diagram, we better demonstrate it with  calculus.
Each q charge exerts force on test charge Q along the axis defined by the force direction, i.e

$F_{Q_j}$ = Const. $Qq_j $

We can choose to express the whole system in 2D co-ordinate system by  complex notation. Assuming test charge Q at the origin, we can say the position of each q charge is as follows,

$q_j  - e^{2ij\pi/13} $

Therefore the total force,

$F_Q$ = Const.  $ Q \sum_{j=1}^{13} q_j$

considering,

$r = e^{2i\pi/13}$

We can see summation is in a geometric progression. We can write it as (putting r)

$F_Q $ = Const. $ Q \sum_{j=1}^{13}$ $r^j$

$F_Q $ = Const. $ (1-r^{13})/(1-r)$

$F_Q $ = Const. $e^{2i\pi 13/13}$

$F_Q$  = 0 

So, the net force experienced by the charge Q is ZERO again.

(EM Griffiths) Problem 2.1 (b)

Question
Suppose one of the 12 q's is removed. What is the force on Q?

Answer

Due to the removal of one q, there is no opposite balancing force existing anymore to nullify the repulsive force of the q sitting at exact opposite corner of vacant site. Rest of the pairs are still in their position, balancing their forces. As a result, the test charge Q will only feel the coulomb force due that single q charge.

If  $F_Q$ is the force experienced by the test charge Q, then

$F_Q = Qq/4\pi\epsilon_or^2$

Where r is the distance between charge Q and q. 

(EM Griffiths) Problem 2.1 (a)

Question

Twelve equal charges, q, are situated at the corners of a regular 12-sided polygon(for instance, one on each numeral of a clock face). What is the net force on a test charge Q at the center?

Ans

As shown in the figure charge Q experiences equal and opposite repulsive forces for every pair of q charges situated at the opposite corners of the given 12-sided polygon. Coulomb force exerted by any q charge upon the test charge Q gets nullified by the other q charge sitting at the exact opposite corner. As a result of that Q experiences ZERO net force as whole.