Electric Field of Uniform Charge Distributions

Here we calculate the electric field due to some uniform charge distributions such as a line of charge, a ring of charge and a uniformly charged disk. You should also notice the symmetry of charge distribution which makes easy to find the electric filed due to that charge distribution.

Electric Field of a Line of Charge

Consider that charge $q$ is distributed uniformly throughout a line as shown in Figure 1 and we are going to calculate the electric field at a point $p$ which lies at a perpendicular distance $x$ from the middle point of the line. It's always easier to start with a coordinate system and the associated symmetry with the charge distribution.

Figure 1 Electric field at a perpendicular distance due to uniform line of charge distribution.

We consider infinitesimal length element of length $dy$ on the line of charge. If the linear charge density is denoted by $\lambda$ which is the charge per unit length, the charge associated within the length $dy$ is $dq =\lambda dy$. The line of charge is divided by the x-axis into equal halves of length $l$, so the total length of the line of charge is $2l$. The charge $dq$ within the length $dy$ is $dq=\lambda dy=\frac{qdy}{2a}$ and the electric field $dE$ due to this charge at point $p$ is

\[dE = k\frac{{dq}}{{{r^2}}} = k\frac{{\lambda dy}}{{({x^2} + {y^2})}} = k\frac{{qdy}}{{2l({x^2} + {y^2})}}\]

Note that $r^2 = x^2 + y^2$ in the above equation. The x-component of the electric field $dE$ is $dE_x=dE\cos \theta $ and the y-component is $dE_y=-dE\sin \theta $ (why negative sign?). Note that the y-component of electric field produced by the upper half part of the line of charge is exactly equal in magnitude and opposite in direction to the y-component of electric field produced by the lower half part of the line of charge and therefore the y-component of the electric field at the point $p$ due to the line of charge is zero. We can understand it by the symmetry consideration of the line of charge at the point $p$. So the total electric filed due to the line of charge at that point is sum of the x-components of electric fields of all charge elements. Since $\cos \theta =\frac{x}{r}=\frac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$, the x-component of $dE$ is,

\[d{E_x} = dE\cos \theta = k\frac{{qxdy}}{{2l{{({x^2} + {y^2})}^{\frac{3}{2}}}}}\]

Now the total electric field at the point $p$ due to the line of charge is determined by integrating $dE_x$ from $-l$ to $l$. Which is

\[{E_x} = E = k\frac{{qx}}{{2l}}\int\limits_{ - l}^{ + l} {\frac{{dy}}{{{{({x^2} + {y^2})}^{\frac{3}{2}}}}}} = k\frac{q}{{x\sqrt {{x^2} + {l^2}} }} \tag{1} \label{1}\]

The $x$ coordinate of the point is constant in the integraion. The unit vector $\hat{i}$ gives the direction along positive x-axis, so the electric field in vector form is

\[\vec E = k\frac{q}{{x\sqrt {{x^2} + {l^2}} }} \hat i \tag{2} \label{2}\]

You can also check that the integration of $d{{E}_{y}}$ from $-l$ to $l$ is zero:

\[d{E_y} = - dE\sin \theta = - k\frac{{qydy}}{{2l{{({x^2} + {y^2})}^{\frac{3}{2}}}}}\]

Now integrating $d{{E}_{y}}$ from $-l$ to $l$,

\[{E_y} = - k\frac{q}{{2l}}\int_{ - l}^l {\frac{{ydy}}{{{{({x^2} + {y^2})}^{\frac{3}{2}}}}}} = 0 \tag{3} \label{3}\]

Only the sum of the x-components of electric fields produced by all elements of the line of charge gives the total electric field at the point $p$.

When the point $p$ is very far from the line of charge that is, the distance $x$ is considerably larger in comparison to the length of the line of charge or symbolically $x >> l$. The term $l^2$ will be much smaller and can be neglected. So the electric field in this case is

\[E = k\frac{q}{{x\sqrt {{x^2}} }} = k\frac{q}{{{x^2}}} \tag{4} \label{4}\]

In this case when the point $p$ is far from the line of charge the entire charge in the line acts as if all the charge were concentrated at the origin of the coordinate system (at the middle point of line of charge)

And when the line of charge is infinitely long, that is the length of the line of charge is very long in comparison to the distance $x$, ($l >> x$), the distance $x$ will be much smaller than $l$ and $x^2$ can be neglected. So, the resulting electric field is

\[E = k\frac{q}{{x\sqrt {{l^2}} }} = k\frac{q}{{lx}} \tag{5} \label{5}\]

As you know the linear charge density is $\lambda =\frac{q}{2l}$ and in terms of the linear charge density, the above expression can be written as,

\[E = k\frac{{2\lambda }}{x} \tag{6} \label{6}\]

So the result is clear that the electric field for the infinitely long line of charge is proportional to $\frac{1}{x}$ instead of $\frac{1}{{{x}^{2}}}$ as for a point charge. And notice that we can not get a real line of charge having infinite length! But when the length of the line of charge is considerably larger in comparison to the distance $x$ and the point $p$ is considerably close enough to the line of charge the electric field is the same as that of an infinite line of charge.

We have also obtained the expression which is same as that of an infinite length line of charge in the electric field calculation using Gauss's Law.

Electric Field of a Ring of Charge

The total charge $q$ is distributed uniformly forming a ring as shown in Figure 2. The charge is uniformly distributed in the ring and so we apply the symmetry to find the total electric field of the ring of charge at a point $p$. In our coordinate system the origin is the centre of the ring and x-axis is perpendicular to the imaginary plane of the ring. The point $p$ where the electric field is being calculated lies at a perpendicular distance $x$ (x-coordinate of point $p$) from the centre of the ring. We divide the entire charge on the ring into small charge elements such as $dq$ and add the electric fields produced by every element at the point $p$.

Figure 2 Electric field due to a ring of charge at a point p. The point $p$ lies at a perpendicular distance $x$ from the imaginary plane of the ring.

Note that each element has both x and y-components of electric field but the y-component is cancelled by the y-component of the opposite charge element in the ring. Therefore the electric field at the point $p$ is only due to the x-component of electric field due to charge elements. In Figure 2 let the radius of the ring be $R$, so the distance between the charge element $dq$ and the point $p$ is $r = \sqrt {{x^2} + {R^2}} $ and the electric field $dE$ at that point due to this charge element is

\[dE = k\frac{{dq}}{{({x^2} + {R^2})}}\]

If you consider the symmetry of the charge distribution you'll find that the y-component of electric field due to one element is cancelled by the the y-component of electric field due to the opposite element in the ring. So the sum of the y-components of electric field due to the entire ring of charge is zero as already noted. Thus the total electric field at the point is the sum of the x-components of electric fields due to all charge elements in the ring. The x-component of the small electric field $dE$ is $d{{E}_{x}}=dE\cos \theta $ and $\cos \theta =\frac{x}{r}$. So,

\[d{E_x} = k\frac{{xdq}}{{{{({x^2} + {R^2})}^{\frac{3}{2}}}}}\]

Now we integrate this expression to get the electric field due to the entire ring of charge. Note that $x$ and $R$ remain constant throughout the whole integration and therefore,

\[{E_x} = E = k\frac{{xq}}{{{{({x^2} + {R^2})}^{\frac{3}{2}}}}} \tag{7} \label{7}\]

If the unit vector $\hat{i}$ is the unit vector along positive x-direction, the electric field in vector form is

\[\vec E = k\frac{{xq}}{{{{({x^2} + {R^2})}^{\frac{3}{2}}}}}\hat i \tag{8} \label{8}\]

If the point $p$ is far enough from the ring that is, $x >> R$. In this case the term ${{R}^{2}}$ can be neglected and the electric field becomes

\[E = k\frac{q}{{{x^2}}} \tag{9} \label{9}\]

Which is the electric field for a charge at a point. It means the entire ring acts as a point charge $q$ if the point $p$ is far enough from the centre of the ring. Now you can easily get the answer for another condition as what if the point $p$ lies at the centre of the ring. For this the value $x$ is zero and the electric field is also zero.

Electric Field of a Uniformly Charged Disk

The Figure 3 shows a disk of total charge $q$. We are going to calculate the electric field due to the disk of charge at a point $p$ which lies at perpendicular distance $x$ from the centre of the disk. The origin of our coordinate system lies at the centre of the disk and the plane of the disk is perpendicular to the x-axis. To get our result we divide the charged disk into small rings of charge $dq$ and get the the electric field at the point $p$ by adding the electric fields produced by every ring element.

Figure 3 Electric field due to a disk of charge. The point $p$ lies at a perpendicular distance $x$ from the plane of the disk.

We already know the electric field of a ring of charge, it's easier to get our result for the disk of charge. As you already know that the total electric field at the point $p$ is only due to the x-component of electric field of ring elements. Consider that the radius of one specific ring is $r$ and thickness is $dr$. So, the ring has area equal to $2\pi rdr$ and the charge $dq$ in the ring is $dq=\sigma (2\pi rdr)$ where $\sigma $ is the charge per unit area. The electric field $dE$ of the ring of charge $dq$ (you already know this) at point $p$ is

\[dE = k\frac{{x\sigma (2\pi rdr)}}{{{{({x^2} + {r^2})}^{\frac{3}{2}}}}}\]

Now the electric field for the entire disk is calculated by integrating the above expression from $0$ to $R$.

\[E = 2k\pi \sigma x\int\limits_0^R {\frac{{rdr}}{{{{({x^2} + {r^2})}^{\frac{3}{2}}}}}} = 2k\pi \sigma x\left( {\frac{1}{x} - \frac{1}{{\sqrt {{x^2} + {R^2}} }}} \right)\]

\[{\rm{or,}}\quad E = 2k\pi \sigma \left( {1 - \frac{1}{{\sqrt {1 + {{\left( {\frac{R}{x}} \right)}^2}} }}} \right) \tag{10} \label{10}\]

In vector form you can write the above expression as

\[\vec E = 2k\pi \sigma \left( {1 - \frac{1}{{\sqrt {1 + {{\left( {\frac{R}{x}} \right)}^2}} }}} \right)\hat i \tag{11} \label{11}\]

where $\hat i$ gives the direction of the electric field vector. Suppose the radius of the disk is large enough in comparison to the distance $x$ so that $R >> x$, the term $1/{\sqrt{1+{{\left( \frac{R}{x} \right)}^{2}}}}$ is negligibly small and can be neglected. And the electric field in this case is

\[E=2k\pi \sigma \tag{12} \label{12}\]

So for the very large disk (or you can think as an infinite disk) the electric field is independent of the distance $x$ and perpendicularly outward from the disk.

Electromagnetism

Was this article helpful?

Physics Key uses third party cookies to show you relevent ads, analyze traffic and for the security. If you continue to use this website, you acknowledge that you have read and understand the Terms and Privacy Policy.