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

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: 

 

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

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

 

Example: wire, circulating ring 

 

2. Surface charge distribution: 

 

$(\sigma)$ - charge per unit Area

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

  

 

Example: plane sheet

3. Volume Charge distribution

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

$
\rho=\frac{Q}{V}=\frac{C}{m^3}=C m^{-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  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 . 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 - 

 

and can be integrated because all electric field lies in the same direction.
Now, here $-d q=\lambda . d x$

$
d E=k_{\mathrm{e}} \frac{d q}{x^2}=k_{\mathrm{e}} \frac{\lambda d x}{x^2}
$

The total lield at $P$ is

$
E=\int_a^{l+a} k_e \lambda \frac{d x}{x^2}
$

If $k_e$ and $\lambda=Q / l$ are constants and can be removed from the integral, then

$
\begin{aligned}
& \qquad 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}
$

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 $\phi_{\text {at the center. }}$

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

$d Q=\lambda R d \theta_{\text {on either side of the horizontal to the arc. }}$
Now resolving the electric field due to small element of the $\operatorname{arc} \lambda R d \theta$,
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 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
$

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{aligned}
& 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{aligned}
$
 

$\begin{aligned} & \quad K=\frac{1}{4 \pi \varepsilon_0} \\ & \therefore \text { Using } \\ & \therefore E=\frac{2 \lambda}{4 \pi \varepsilon_0 R} \sin \frac{\phi}{2}\end{aligned}$