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Electric Field Intensity: Continuous Charge Distribution - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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8 Questions around this concept.

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MEDIUM

A wire, of length L(=20 cm), is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges $\pm Q$ , $\left | Q \right |=10^{3} \varepsilon _{0}$ Coulomb where $\varepsilon _{0}$ is the permittivity (in SI units) of free space] the net electric field at the centre O of the semi-circular arc would be :

$\left(50 \times 10^3 N / C\right) \hat{j}$

$\left(25 \times 10^3 N / C\right) \hat{i}$

$\left(25 \times 10^3 N / C\right) \hat{j}$

$\left(50 \times 10^3 N / C\right) \hat{i}$

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A point charge Q is placed outside a hollow conductor of radius R, at a distance $r(r>R)$ from its centre c. The electric field at the centre c of the conductor due to charges induced on it.

$\text { prop to } r^2$

$\text { prop to } r^{-2}$

$\text { Prop. to\, \,r }$

$\text { Prop to } r^3$

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Concepts Covered - 1

Electric field due to continuous charge distribution

Discrete charge distribution: A system consisting of many individual charges.

Continuous charge distribution: An amount of charge distributed uniformly or non uniformly on a body. It is of three types -

1. Linear charge distribution:

$\left ( \lambda \right ) -$ charge per unit length.

$\dpi{100} \lambda=\frac{q}{L}=\frac{C}{m}=Cm^{-1}$

Example: wire, circulating ring

2. Surface charge distribution:

$\left ( \sigma \right )-$ charge per unit Area

$\dpi{100} \sigma=\frac{Q}{A}=\frac{C}{m^{2}}=Cm^{-2}$

Example: plane sheet

3. Volume Charge distribution

$\left ( \rho \right )-$ charge per unit volume.

$\dpi{100} \rho=\frac{Q}{V}=\frac{C}{m^{3}}=Cm^{-3}$

Example - charge on a dielectric sphere etc.

Now we will discuss one example and derivation of the electric field due to uniformly charged rod.

So, let us consider a rod of length $l$ which has uniformly positive charge per unit length lying on x-axis, dx is the length of one small section. This rod is having a total charge Q and dq is the charge on dx segment. The charge per unit length of the rod is $\lambda$ . We have to calculate the electric field at a point P which is located along the axis of the rod at a distance of 'a' from the nearest end of Rod as shown in the figure -

The field $\vec{dE}$ at P due to each segment of charge on the rod is in the negative "x" direction because every segment of the rod carry positive charge. In this every segment of the rod is producing electric field in the negative "x" direction, so the sum of electric field can be added directly and can be integrated because all electric field lies in the same direction.

Now, here - $dq = \lambda.dx$

$\begin{array}{l}{\qquad d E=k_{\mathrm{e}} \frac{d q}{x^{2}}=k_{\mathrm{e}} \frac{\lambda d x}{x^{2}}} \\ \\{\text { The total lield at } P \text { is }} \\ \\ {\qquad E=\int_{a}^{l+a} k_{e} \lambda \frac{d x}{x^{2}}} \\ \\ {\text { If } k_{e} \text { and } \lambda=Q/l \text { are constants and can be removed from the }} {\text { integral, then }} \\ \\ {\qquad \begin{aligned} E &=k_{e} \lambda \int_{a}^{l+a} \frac{d x}{x^{2}}=k_{e} \lambda\left[-\frac{1}{x}\right]_{a}^{l+a} \\ \\ \Rightarrow k_{e} \frac{Q}{l}\left(\frac{1}{a}-\frac{1}{l+a}\right)=\frac{k_{e} Q}{a(l+a)} \end{aligned}}\end{array}$

Now if we slide the rod toward the origin and the $a\rightarrow 0$ , then due to that end, the electric field is infinite.

Electric field strength due to a charged circular arc at its centre

Let's try to find out the electric field at the center of an arc of linear charge density $\lambda$ , radius R
subtending angle $\phi$ at the center.

If Q is the total charge contained in the arc then,
$\lambda=\frac{Q}{R \phi}$

By symmetry, we know that the electric field due to the arc will be radially outward at the center.

Now consider a small element of the arc of charge
$d Q=\lambda R d \theta$ on either side of the horizontal to the arc.

Now resolving the electric field due to small element of the arc $\lambda R d \theta$ ,

we see that both the vertical components get canceled and all that remains will be the horizontal component of the electric field due to the corresponding small
elements.

So we have $2 d E \cos \theta$ due to all such corresponding small elements on the arc,

so, by integrating all those electric fields due to these small elements we get the electric field due to the whole arc.

$E=\int_{0}^{\frac{\phi}{2}} 2 d E \cos \theta \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1)$
We know that electric due to charge dQ is given by,
$d E=K \frac{d Q}{R^{2}}$
But, $d Q=\lambda R d \theta$

$d E=K \frac{\lambda R d \theta}{R^{2}}=K \frac{\lambda d \theta}{R}$

Substituting (2) in (1)

$\begin{array}{l} E=\int_{0}^{\frac{\phi}{2}} 2 K \frac{\lambda \cos \theta d \theta}{R} \\ \Rightarrow E=\frac{2 K \lambda}{R} \int_{0}^{\frac{\phi}{2}} \cos \theta d \theta \\ \Rightarrow E=\frac{2 K \lambda}{R}[\sin \theta]_{0}^{\frac{\phi}{2}} \\ \Rightarrow E=\frac{2 K \lambda}{R}\left(\sin \left(\frac{\phi}{2}\right)-\sin 0\right) \\ \Rightarrow E=\frac{2 K \lambda \sin \frac{\phi}{2}}{R} \end{array}$
Using $K=\frac{1}{4 \pi \varepsilon_{0}}$
$\therefore E=\frac{2 \lambda}{4 \pi \varepsilon_{0} R} \sin \frac{\phi}{2}$