Question

22-52ashows a non-conducting rod with a uniformly distributed charge Q. The rod forms a half-circle with radius Rand produces an electric field of magnitudeat its center of curvature P. If the arc is collapsed to a point at distance Rfrom P(Fig. 22-52b), by what factor is the magnitude of the electric field at Pmultiplied?

Short answer

Answer:

If the arc is collapsed to a point at distance R from P, the magnitude of the electric field at P is multiplied by 1.57.

Step-by-step solution

The given data

A non-conducting rod of uniform charge,Q , forms a half circle with radius R and produces an electric field,$E_{arc}$ at its center of curvature.

Understanding the concept of electric field 

Using the concept of the electric field of a charged rod, we can get the net electric field of the distribution at the center of the formed circle.

Formulae:

“Electric field of a charged circular rod,” we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc, (i)

where, r = radius of the circle

λ = charge per unit length of the rod

The length of an arc of a circle,$L=r^0(0inradians)$ (ii)

The linear density of a distribution,$\lambda =q\backslash L$ (iii)

The electric field of a particle, (iv)

Calculation of the multiplying factor of the magnitude of the electric field 

“Electric field of a charged circular rod,” we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc using equation (i), and the given data is given as:

Now, using equations (ii) and (iii) in the above equation, we can get the electric field value at the arc as follows:

Now, we can get the multiple factors of the electric field of the particle in comparison to the electric field at arc by dividing equation (iv) by equation (a) as follows:

With given the angle for half circle, , we have the required ratio as:

Hence, the value of the multiplying factor is 1.57