Q28P - Fundamentals Of Physics

Question

Charge is uniformly distributed around a ring of radius R=2.40cm, and the resulting electric field magnitude E is measured along the ring’s central axis (perpendicular to the plane of the ring). At what distance from the ring’s center is E maximum?

Step-by-Step Solution

Answer

The distance from the ring’s center at which the electric field has maximum value is 1.70 cm.

Step 1: The given data

• Radius of the circular ring, R =2.40 cm.
• The electric field is perpendicular to the surface of the ring, that is, along the central axis.

Step 2: Understanding the concept of electric field

Using the concept of the electric field due to the ring, we can get the value of the distance at which the electric field is found to be maximum by differentiating the given equation of the electric field.

Formula:

Electric field due to ring, $E=\frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{qz}{{\left(z^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}$                                                       (i)

Where R = radius of the ring.

          z = distance of the field point on the axis of the ring.

Step 3: Calculation of the position of the maximum magnitude

The electric in this case can be given using equation (i). Again, we can find the position of the maximum electric field by differentiating equation (i) for z and setting the result equal to zero as follows:

$$\begin{gathered} \frac{d}{dx}\left[\frac{1}{4{\mathrm{\pi \epsilon }}_0} \frac{qz}{{\left(z^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}\right]=0 \\ \frac{q}{4\mathrm{\pi \epsilon }}\frac{R^2-2z^2}{{\left(z^2+R^2\right)}^{{\displaystyle \frac{5}{2}}}}=0 \\ z=R/\sqrt{2} \\ =2.40 \mathrm{cm}/\sqrt{2} \\ =1.70 \mathrm{cm} \end{gathered}$$

Hence, the value of the required position of maximum magnitude is 1.70 cm