Question

A Semicircular rod is charged uniformly with a total charge \(\mathrm{Q}\) coulomb. The electric field intensity at the centre of curvature is $\ldots \ldots \ldots$ (A) \(\left[(2 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (B) \(\left[(3 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (C) \(\left[(\mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (D) \(\left[(4 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\)

Short answer

The short answer is: The electric field intensity at the center of curvature of a uniformly charged semicircular rod is given by: \[\boxed{E = \frac{kQ}{R^{2}}}\] The correct answer is (C).

Step-by-step solution

Define the given values

A semicircular rod is uniformly charged with a total charge Q. Let's denote the radius of the semicircular rod as R.

Divide the rod into small segments

To find the electric field intensity at the center of curvature, divide the semicircular rod into n small charge segments \(dq\). Each segment will have an angle \(d\theta\), where \(\theta\) ranges from 0 to \(\pi\).

Find the electric field (dE) produced by each segment

The electric field produced by each charge segment \(dq\) at the center of curvature can be calculated using Coulomb's law: \[dE = \frac{kdq}{R^2}\] where k is the Coulomb's constant, and \(R^2\) is the square of the distance from the center.

Express dq in terms of dθ

As the charge distribution is uniform, the charge per unit angle will be constant. So, we can write: \[\frac{dq}{d\theta} = \frac{Q}{\pi}\] Now, we can express \(dq\) in terms of \(d\theta\) as: \[dq = \frac{Q}{\pi}d\theta\]

Substitute dq in the expression for dE

Using the expression for \(dq\) in terms of \(d\theta\), put the value of \(dq\) in the equation for \(dE\): \[dE = \frac{k\left(\frac{Q}{\pi}d\theta\right)}{R^2}\] This simplifies to: \[dE = \frac{kQd\theta}{\pi R^2}\]

Calculate the net electric field (E)

To find the net electric field intensity at the center of the curvature, integrate \(dE\) over the semicircular rod: \[E = \int_{0}^{\pi} \frac{kQd\theta}{\pi R^2}\] This integral results in: \[E = \frac{kQ}{\pi R^2} \int_{0}^{\pi} d\theta\] \[E= \frac{kQ}{\pi R^2} (\pi) \] Finally, we get: \[E = \frac{kQ}{R^{2}}\]

Compare the result with the given options

The calculated electric field intensity is: \[E = \frac{kQ}{R^2}\] Comparing it with the given options, we see that it matches option (C): \[\left(\frac{KQ}{\pi R^{2}}\right)\] So the correct answer is (C).