Consider a charge q, placed on each of the comers of a cube of Side length s. Sh

Question

Consider a charge q, placed on each of the comers of a cube of Side length s. Show that the magnitude of the electric field a, the center of any face of the cube has of 2.18 kq/s^2. Estimate the value of the electric field at point P in the figure due to the rod of length L and total uniformly distributed charge Q (a) Do this by dividing the charged rod into 4 equal length segments and treating each segment as a point charge at the center of each segment (b) How could you improve on your estimate? Reading a journal from that crazy old retired physics professor on the hill, you stumble upon a scheme to generate high frequency (HF) radio waves (lambda = 10 m). It requires generating an electric field that diverges from a point and increases in strength linearly with respect to distance from that point. The sketch from that crazy old nun's book is to the right, (a) What type of common charge would create an oscillation if placed in this field? Explain why the others would not. at rest, the divergence point and still oscillate? c) What must the E-field strength coefficient to generate the HF radio waves? Extra credit (and possible greater accolades) - How could such a field be generated?

Explanation / Answer

Each side of cube has length a (say)

At each vertex there is a charge q

We need to find Electric field at middle of any face of the cube.

There are four charges(at vertices on same face as the face on which we measure field) at a distance of a/Sqrt2. The elecric field due to these four charges is 0 at middle of face ,since the charges at opposite vertices in this this square cancel each other's field

The charge at other four corners are at a distance of sqrt ( a^2 + a^2/2) = sqrt(3/2) a,from the location where we measure field

The feild at middle of face due to each charge q on opposite face is

E1 = kq/{sqrt(3/2) a}^2 = (2/3) kq/a^2

Now the horizontal component of E1

E(h) = E1 cos(x) = (2/3) kq/a^2 X sqrt(2/3)

Now the hiorizontal component of field due to all four charges gets added up. The vertical components field cancel each other.

E = 4E(h) = (8/3) sqrt( 2/3) kq/a^2 = 2.18 kq^2/a^2

(The feild at midddle of a side is 4 times the field due to any one of the charges at corner of opposite face.)

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