Electric Field due to Uniformly Charged Ring

Electric Field due to Uniformly Charged Ring

1 Star2 Stars3 Stars4 Stars5 Stars (5 votes, average: 4.80 out of 5)
Loading...

Rate This Article With 5 Stars.

Last updated on August 29th, 2019 at 04:00 pm

The main objective of this article is to calculate the Electric Field due to a Uniformly Charged Ring at a Point on its Axis. Prerequisite of this calculation is the knowledge of Electric Field due to Point Charge and Line Charge Distribution. We recommend to visit those article before continuing. Moreover, you can browse another theoretically similar article on Gravitational Field due to a Uniform Ring at a Point on its Axis for a comparative study.

Electric Field due to Uniformly Charged Ring at a Point on its Axis

Here we will calculate the electric field at an axial point P due to a uniformly charged ring of charge +Q. The center of the ring is at point O. From the figure shown below, the circumference of the ring makes an angle \theta with the line OP drawn from the center of the ring to the test point P. We will consider a very small charge element dq (shown as red charge element), on the circumference of the ring. The center of this charge element is at point A. Now to calculate the field at point P due to this charge element, we will use the expression of the electric field derived in our previous article, Electric Field due to Point Charge.

    \[dE=\frac{dq}{4 \pi \epsilon_0 z^2} \]

electric field due to uniformly charged ring

The direction of this field is along \vec{AP}. Now, we can resolve this field vector into two components, one along the positive x-direction and another along the negative y-direction. This is shown as the red vectors in the diagram above. The x component of the electric field due to the charge element dq is

Advertisement

    \[dE_x=dEcos(\theta)=\frac{dq}{4\pi \epsilon_0 z^2}cos(\theta) \]

The y component of the electric field due to the charge element dq is

    \[dE_y=dEsin(\theta)=\frac{dq}{4\pi \epsilon_0 z^2}sin(\theta) \]

To calculate the total electric field at point P due to this circular charged ring, we have to integrate dE over the whole ring. However, due to the axial symmetry of the total system, the y component will vanish and we will have the x component only. What does it mean? In order to explain this, we have considered another charge element dq' (shown in green). Electric field due to this charge element is shown as a green vector. Horizontal and vertical components are also shown in green. Now, for each dq charge there will be another dq positioned at a radially opposite point on the circumference of the charged ring. The y component of the electric field due to these two charges cancel each other, whereas the x components add up. This is true for each and every point on the charged circular ring. Thus we are left with x component of the field only.

    \[E=\int{\frac{dq}{4\pi \epsilon_0 z^2}cos(\theta)} \]

    \[=\frac{cos(\theta)}{4 \pi \epsilon_0 z^2}\int{dq} \]

    \[=\frac{Q cos(\theta)}{4 \pi \epsilon_0 z^2} \]

    \[E=\frac{Q r}{4\pi \epsilon_0 (a^2+r^2)^{\frac{3}{2}}}\]

Or

    \[\boxed{\vec{E}=\frac{Qr}{4\pi\epsilon_0 (a^2+r^2)^{\frac{3}{2}}}\hat{r}} \]

The direction of the field along \vec{OP}. This is the final expression for the electric field due to uniformly charged ring at a point on its axis. In the next article, we will use this expression to calculate the Electric Field due to a Uniformly Charged Disc. Can you solve the problem for an off-axis point?

Advertisement