Question

[A] To determine the electric field at the focus of a parabolic segment of charge, the simplest approach to calculate the answer is to use Colomb

Answer 1

A)

The advantages of using this method is
1) By symmetry, only one component of electric field is non-zero
We can split a line element into sine and cosine components. If we take two such line elements symmetrically placed in the y axis, then the cosine components will add up and the sine components will cancel out so the integration will be easy.
2) We can not apply Gauss's law here.
It is very difficult to construct a Gaussian surface at the point we want to calculate the electric field.

Example: Electric field at the center of circular arc carrying a charge Q
Take the radius of the arc to be R, length, L and theta is the angle subtends with y-axis
If N is the linear charge density, dq = N dL, dq is the element of charge occupies in the element of length dL
Take the constant 1/(4 pi x epsilon naught) = k
We can use the symmetry of the circular arc to calculate the electric field. From the center of the circular arc, consider two points equidistant from the center. The y components of the electric fields adds up and the x components cancels out. Here theta is the angle subtended to y axis.


  

B)

Step 1: Calculate the work done in bringing each individual charges.
1) Work done in bringing q1 from infinity
W1 = 0 since there is no charge to attract or repel q1.
2)Work done in bringing q2 from infinity.
There is only an attractive force from q1
?W2 = k (-q1 x q2)/d
?3)Work done in bringing q3 from infinity.
?There is attraction from q2 and repelsion from q1
?Here the distance between q1 and q3 is 1.414 d.
?W3 = k(-q1x -q3) / (1.414d) + k(-q3 x q2)/d.
4) work done in bringing q4 from infinity.
There is attraction from q1, repulsion fom q2 and attraction from q3
?q1 and q3 are are at a distance d from q4 and q2 is at a distance of 1.414 d from q2
W4 = k(q4 x -q3) / d + k (q4 x q2) / 1.414d + k(q4 x -q1)/d.

Step 2: Add all the work to get the potential energy
Here q1 = q2 = q3 = q4 = q
W = W1 + W2 + W3 + W4
? 0 + [-kq²/d] + [-kq²/d +kq²/1.414d] + [-kq²/d + kq²/1.414d - kq²/d]
   = -4kq²/d + 2kq²/1.414d = - 2.58 kq²/d.

The energy is negative in the sense there is no work done on the system, ie for assembling this configuration , no external force is required and we need exactly this much energy to de disassemble this system