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Chapter 11: Problem 1544

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

Expert verified
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

01

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.
02

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 \(\pi\).
03

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.
04

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: \[dq = \frac{Q}{\pi}d\theta\]
05

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: \[dE = \frac{kQd\theta}{\pi R^2}\]
06

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: \[E = \frac{kQ}{R^{2}}\]
07

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).

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Most popular questions from this chapter

A parallel plate air capacitor has a capacitance \(18 \mu \mathrm{F}\). If the distance between the plates is tripled and a dielectric medium is introduced, the capacitance becomes \(72 \mu \mathrm{F}\). The dielectric constant of the medium is (A) 4 (B) 12 (C) 9 (D) 2

A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) $\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]$ (B) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]$ (C) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]$ (D) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$

The displacement of a charge \(Q\) in the electric field $E^{-}=e_{1} i \wedge+e_{2} j \wedge+e_{3} k \wedge\( is \)r^{-}=a i \wedge+b j \wedge$ The work done is \(\ldots \ldots\) (A) \(Q\left(e_{1}+e_{2}\right) \sqrt{\left(a^{2}+b^{2}\right)}\) (B) \(Q\left[\sqrt{ \left.\left(e_{1}^{2}+e_{2}^{2}\right)\right](a+b)}\right.\) (C) \(Q\left(a e_{1}+b e_{2}\right)\) (D) \(\left.Q \sqrt{[}\left(a e_{1}\right)^{2}+\left(b e_{2}\right)^{2}\right]\)

Two point charges of \(+16 \mathrm{c}\) and \(-9 \mathrm{c}\) are placed $8 \mathrm{~cm}\( apart in air \)\ldots \ldots\(.. distance of a point from \)-9$ c charge at which the resultant electric field is zero. (A) \(24 \mathrm{~cm}\) (B) \(9 \mathrm{~cm}\) (C) \(16 \mathrm{~cm}\) (D) \(35 \mathrm{~cm}\)

The potential at a point \(\mathrm{x}\) (measured in \(\mu \mathrm{m}\) ) due to some charges situated on the \(\mathrm{x}\) -axis is given by \(\mathrm{V}(\mathrm{x})=\left[(20) /\left(\mathrm{x}^{2}-4\right)\right]\) Volt. The electric field at \(\mathrm{x}=4 \mu \mathrm{m}\) is given by (A) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (B) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (C) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (D) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction
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