Question

Derive the expression for net Electrostatic Field Intensity at a distance "x" on the geometrical axis of a circular ring of radius R. The ring has been given a uniform charge Q.

#Revision Q8
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Answer 1

Answer:

force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.

Answer 2

$\underline{\underline{\sf{\maltese\:\:Question}}}$

• Derive the expression for net Electrostatic Field Intensity at a distance $\sf{x}$ on the geometrical axis of a circular ring of radius $\sf{R}$. The ring has been given a uniform charge $\sf{Q}$

$\underline{\underline{\sf{\maltese\:\:Given}}}$

• The circular ring is having the uniform charge $\sf{Q}$

• The distance on the geometrical axis of a circular ring is $\sf{x}$

• The circular ring is having radius $\sf{R}$

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

• The expression for net Electrostatic Field Intensity at a distance $\sf{x}$ on the geometrical axis of a circular ring of radius $\sf{R}$ is

$\boxed{\boxed{\sf{\displaystyle\frac{Qx}{4\pi\varepsilon_0\left(x^2+R^2\right)^{\left(3/2\right)}}}}}$

$\underline{\underline{\sf{\maltese\:\:Explanation}}}$

The diagram of the circular ring having element charge can be drawn as, (Kindly Refer to Attachment - 1)

$\textsc{Points Regarding The Diagram,}$

• The circular ring carrying total charge $\sf{Q}$ with a radius of $\sf{R}$.

• The electric field can be represented as $\overrightarrow{\sf{dE}}$ at point A on the central $\sf{x\:-\:axis}$

• The elemental strip is taken having the small charge $\overrightarrow{\sf{dQ}}$

• The component of the electric field along the x-axis direction can be represented by $\overrightarrow{\sf{dE_x}}$

• The perpendicular component of the electric field can be shown as $\overrightarrow{\sf{dE_\perp}}$

Now, considering an element of the charge $\sf{dQ_1}$. So there must be an element of equal charge $\sf{dQ_2}$ in the direction opposite to that charge.

• The diagram of the circular ring having equal and opposite charge can drawn as, (Kindly Refer to Attachment - 2)

$\textsc{Points Regarding The Diagram,}$

• The perpendicular components of the electric field 1 and 2 will cancel out with each other.

• So, this figure represents the symmetric representation of the distribution of the charges.

• Therefore the net electric field of the full circular ring must be in the same line of $\sf{x}$ - axis.

• The distance from point A to $\sf{dQ_1}$ is represented as $\sf{a}$

• The net summation of the charges perpendicular to the $\sf{x}$ - axis direction is zero as they are in opposite direction to each other.

• Here, $\sf{dE_1}$ and $\sf{dE_2}$ are the electric fields for the charges $\sf{dQ_1}$ and $\sf{dQ_2}$                                                              

• Here, $\sf{dE_{1x}}$ and $\sf{dE_{2x}}$ are the components of electric fields of 1 and 2 along $\sf{x}$ - axis and $\sf{d E_{1 \perp}}$ and $\sf{d E_{2 \perp}}$ are the components of electric field in perpendicular direction.

So, the component of the electric field is along the x-axis because to the small element can be calculated as,

$\sf{\displaystyle d E_{x}=\frac{k \times d Q}{a^{2}} \cos \phi \quad \ldots \ldots(1)}$

• Here, k is the constant whose value is $\sf{1 /\left(4 \pi \varepsilon_{0}\right)}$

• The radius of the circular ring can be expressed from the figure as $\sf{a^{2}=x^{2}+R^{2}}$

The angle can be calculated as,

$\to\sf{\displaystyle\cos \phi=\frac{x}{a}}$

$\to\sf{\displaystyle\cos \phi=\frac{x}{\sqrt{a}^2}}$

$\to\sf{\displaystyle\cos \phi=\frac{x}{\sqrt{x^{2}+R^{2}}}}$

• Substituting the values of radius and angle in the above equation (1)

$\sf{\displaystyle d E_{x}=\frac{k \times d Q}{a^{2}}\left(\frac{x}{a}\right)}$

$\to\sf{\displaystyle d E_{x}=\frac{k \times d Q \times x}{\left(\left(x^{2}+R^{2}\right)^{3 / 2}\right)}}$

The net electric field because of the total charge in the x direction can be calculated as,

$\sf{\displaystyle E_{x}=\int \frac{k x \times d Q}{\left(x^{2}+R^{2}\right)^{(3 / 2)}}}$

Substituting the value as $\sf{1 /\left(4 \pi \varepsilon_{0}\right)}$ for k in the above equation and rearranging the equation.

$\to\sf{\displaystyle E_{x}=\frac{x}{4 \pi \varepsilon_{0}\left(x^{2}+R^{2}\right)^{(3 / 2)}} \int d Q}$

• Here, $\sf{\varepsilon_{0}}$ is the permittivity of free space.

After integration the equation expressed as,

$\to\sf{\displaystyle E_{x}=\frac{Q x}{4 \pi \varepsilon_{0}\left(x^{2}+R^{2}\right)^{(3 / 2)}}}$

• Here, $\sf{Q}$ is the total charge on the circular ring.

Therefore, the expression for net Electrostatic Field Intensity at a distance $\sf{x}$ on the geometrical axis of a circular ring of radius $\sf{R}$ is

$\boxed{\boxed{\sf{\displaystyle\frac{Qx}{4\pi\varepsilon_0\left(x^2+R^2\right)^{\left(3/2\right)}}}}}$