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Chapter 20: Problem 8

The ring in Example 20.6 carries total charge \(Q,\) and the point \(P\) is the same distance \(r=\sqrt{x^{2}+a^{2}}\) from all parts of the ring. So why isn't the electric field of the ring just \(k Q / r^{2} ?\)

Short Answer

Expert verified
The electric field of the ring is not simply \(k Q / r^{2}\) because the total field at point \(P\) is the result of a vector sum of the electric field contributions from all infinitesimal charge elements on the ring, where the tangential components of the field vectors cancel each out and only the radial components contribute to the resultant field.

Step by step solution

01

Understanding the physical scenario

The point P, which is at an equal distance from all sections of the ring, experiences an electric field due to every infinitesimally small charge distributed along the circumference of the ring. Each of these electric fields can be broken into two components - one along the axial line (radial component) and the other perpendicular to it, in the plane of the ring (tangential component).
02

Vector summation of the electric fields

The cylindrical symmetry of the configuration implies that the tangential components from the diametrically opposite charges on the ring cancel out each other, leaving behind only the radial components to contribute to the net electric field. Therefore, the electric field at point P is not simply resultant from the direct product of the total charge and the square of distance.
03

Calculation of the resultant electric field

The net electric field at point P can be computed as the vector sum of the electric field vectors originating from each infinitesimally small charge \(dq\) on the charge ring and it should sum over the whole charge \(Q\). This is given by the expression: \(E = \int_{ring} \frac{k* dq * \cos(\theta)}{r^{2}}\), where \(dq\) is an infinitesimal charge element, \(\theta\) is the angle made by the electric field vector with the axial line, and \(r\) is the distance of \(P\) from \(dq\).

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