Q29P

Question


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



Step-by-Step Solution

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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.

1The given data

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

2Understanding 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=r0(0 in radians)                                 (ii)

The linear density of a distribution,λ=q\L                               (iii)

The electric field of a particle,    (iv)

 




3Calculation 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