A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d

Question

A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d<<R). The electric field at the centre of the ring will now bea)directed towards the gap, inversely proportional to R3b)directed towards the gap, inversely proportional to R2c)directed away from the gap, inversely proportional to R3d)directed away from the gap, inversely proportional to R2.Correct answer is option 'A'. Can you explain this answer?

Answer

Since the ring has uniform charge distribution, we can treat it as a collection of infinitesimally small charged elements. Let's consider an element of length $dl$ on the ring, and let its position be denoted by the angle $ heta$ (measured from some reference point on the ring). The charge on this element is then:

$$dq = \frac{Q}{2\pi R} dl$$

where $\frac{Q}{2\pi R}$ is the charge density of the ring.

Now, let's consider the electric field due to this element at a point P located a distance $x$ from the center of the ring, as shown in the diagram below:

![Ring Diagram](https://i.imgur.com/7X9Oemc.png)

The electric field due to this element can be found using Coulomb's law:

$$dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2}$$

where $r$ is the distance from the element to point P. Using the diagram, we can see that:

$$r^2 = x^2 + R^2 - 2xR\cos heta$$

Substituting in the expressions for $dq$ and $r^2$ and integrating over the length of the ring, we get:

$$E = \frac{1}{4\pi\epsilon_0} \int_{ heta_1}^{ heta_2} \frac{Q}{2\pi R} \frac{dl}{(x^2 + R^2 - 2xR\cos heta)^{3/2}}$$

where $ heta_1$ and $ heta_2$ are the angles corresponding to the beginning and end of the removed section of the ring.

To evaluate this integral, we can make use of the substitution $u = \cos heta$, $du = -\sin heta d heta$, and the fact that $dl = Rd heta$. This gives:

$$E = -\frac{1}{4\pi\epsilon_0} \frac{Q}{2\pi R} \int_{-1}^{a} \frac{Rd heta}{(x^2 + R^2 - 2xRu)^{3/2}}$$

where $a = \cos( heta_1)$, and we've used the fact that $\cos( heta_2) = \cos( heta_1 + d) = \cos( heta_1)$, since $d$ is small.

Making the substitution $t = x/Ru$, we get:

$$E = -\frac{1}{4\pi\epsilon_0} \frac{Q}{2\pi R^2} \int_{x/R}^{x/(R\cos heta_1)} \frac{dt}{(1 + t^2 - 2t)^{3/2}}$$

Simplifying the denominator using the identity $1 - 2t + t^2 = (1-t)^2$, we get:

$$E = -\frac{1}{4\pi\epsilon_0} \frac{Q}{2\pi R^2} \int_{x/R}^{x/(R\cos heta_1)} \frac{dt}{

Similar Class 12 Doubts

The figure shows a nonconducting ring which has positive and negative charge non uniformly distributed on it such that the total charge is zero. Which of the following statements is true ?

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