Electric Field Strength Due To A Uniformly Charged Rod At A General Point

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Electric Field Strength due to a Uniformly Charged Rod at a General Point

To find electric field strength at any general point P, let the point P be at a perpendicular distance R from the rod and let Q be the total charge (For specificity, Q is taken as positive, but the results apply to either sign of charge on the rod) on the rod. Let the linear charge density of the rod is λ (=Q/L).

The right end of the rod is located at angle θ_R measured counter-clockwise from the y axis and the left end of the rod is located at angle θ_L measured clockwise from the y axis. Using sign convention θ_R will be taken as positive and θ_L will be taken as negative.

To find electric field strength at P, we consider an element on rod of length dx at a distance x from point O as shown in figure. Now if dE is the electric field at P due to the element, then

$\displaystyle dE=\frac{kdq}{{{r}^{2}}}$ , here $\displaystyle dq=\frac{Q}{L}dx$ = λdx

The x component of the electric field is seen from this diagram will be :-

$\displaystyle {{E}_{x}}=-\int{\frac{kdq}{{{r}^{2}}}}\sin \theta =-\int{\frac{k\lambda dx}{{{r}^{2}}}}\sin \theta$ …..(1)

As dq = λdx

[ Negative sign is because dEsinθ is in negative x-direction ]

and y component :-

$\displaystyle {{E}_{y}}=\int{\frac{kdq}{{{r}^{2}}}}\cos \theta =\int{\frac{k\lambda dx}{{{r}^{2}}}}\cos \theta$ …..(2)

Now from above figure x = R tanθ ⇒ dx = R sec²θ dθ …..(3)

and r² = R² + x²= R² + (R tanθ)² = R² (1 + tan²θ) = R²sec²θ …..(4)

Using the values of equations (3) and (4), in equations (1) and (2), we get

$\displaystyle {{E}_{x}}=-\int{\frac{k\lambda (R{{\sec }^{2}}\theta d\theta )}{{{R}^{2}}{{\sec }^{2}}\theta }}\sin \theta =-\frac{k\lambda }{R}\int\limits_{{{\theta }_{L}}}^{{{\theta }_{R}}}{\sin \theta d\theta }$

⇒ $\displaystyle {{E}_{x}}=-\frac{k\lambda }{R}(-\cos \theta )_{{{\theta }_{L}}}^{{{\theta }_{R}}}=\frac{k\lambda }{R}(\cos {{\theta }_{R}}-\cos {{\theta }_{L}})$ …..(5)

and similarly…

$\displaystyle {{E}_{y}}=\int{\frac{k\lambda (R{{\sec }^{2}}\theta d\theta )}{{{R}^{2}}{{\sec }^{2}}\theta }}\cos \theta =\frac{k\lambda }{R}\int\limits_{{{\theta }_{L}}}^{{{\theta }_{R}}}{\cos \theta d\theta }$

⇒ $\displaystyle {{E}_{y}}=\frac{k\lambda }{R}(sin\theta )_{{{\theta }_{L}}}^{{{\theta }_{R}}}=\frac{k\lambda }{R}(sin{{\theta }_{R}}-\sin {{\theta }_{L}})$ …..(6)

Note :- Equation (5) and (6) are the general results. The sign convention for θ is not applied till now. We will use these results in various interesting cases with sign convention for θ. Here are the different cases…..

Case I :- To find Electric Field Strength when the point P lies on the perpendicular bisector of the rod

On the perpendicular bisector of a the finite rod, θ_R = +θ and θ_L = -θ.

Then From Eq. (5), E_x = 0, and Eq. (6) becomes

$\displaystyle {{E}_{y}}=\frac{k\lambda }{R}[sin(+\theta )-\sin (-\theta )]=\frac{k\lambda }{R}(2\sin \theta )$

⇒ $\displaystyle {{E}_{y}}=\frac{2k\lambda }{R}\frac{L/2}{\sqrt{{{R}^{2}}+{{(L/2)}^{2}}}}$

Or $\displaystyle {{E}_{y}}=\frac{kQ}{R\sqrt{{{R}^{2}}+{{L}^{2}}/4}}$ …..(7)

In this case also if R >>> L, then the result becomes $\displaystyle {{E}_{y}}=\frac{kQ}{{{R}^{2}}}$ , this is nothing but Electric Filed Intensity due to a point charge.

Case II :- To find Electric Field Strength for an infinitely long rod

When the rod is of infinite length, then θ_R = +90° and θ_L = -90°.

From Eq. (5), E_x = 0, and Eq. (6) gives

$\displaystyle {{E}_{y}}=\frac{k\lambda }{R}[sin(+90)-\sin (-90)]=\frac{k\lambda }{R}(1+1)$

⇒ $\displaystyle {{E}_{y}}=\frac{2k\lambda }{R}$ …..(8)

Case III :- To find Electric Field Strength when the point P lies exactly above the right end of the finite rod

In this case θ_R = +0° and θ_L = -θ°(Let).

Then equation (5) gives…

$\displaystyle {{E}_{x}}=\frac{k\lambda }{R}[(\cos 0-\cos (-\theta )]=\frac{k\lambda }{R}(1-\cos \theta )$

⇒ $\displaystyle {{E}_{x}}=\frac{k\lambda }{R}\left[ 1-\frac{R}{\sqrt{{{R}^{2}}+{{L}^{2}}}} \right]$ …..(9)

And equation (6) gives…

$\displaystyle {{E}_{y}}=\frac{k\lambda }{R}[sin0-\sin (-\theta )]=\frac{k\lambda }{R}\sin \theta$

⇒ $\displaystyle {{E}_{y}}=\frac{k\lambda }{R}\frac{L}{\sqrt{{{R}^{2}}+{{L}^{2}}}}$ …..(10)

Case IV :- To find Electric Field Strength when the point P lies exactly above the right end of a semi- infinite rod

In that case θ_R = +0° and θ_L = -90°.

Equation (5) gives…

$\displaystyle {{E}_{x}}=\frac{k\lambda }{R}[(\cos 0-\cos (-90)]=\frac{k\lambda }{R}(1-0)$

⇒ $\displaystyle {{E}_{x}}=\frac{k\lambda }{R}$ …..(11)

And from equation (6)…

$\displaystyle {{E}_{y}}=\frac{k\lambda }{R}[sin0-\sin (-90)]=\frac{k\lambda }{R}(0+1)$

⇒ $\displaystyle {{E}_{y}}=\frac{k\lambda }{R}$ …..(12)

Note that equations (11) and (12) gives E_x = E_y, which means that…

“Electric field due to a charged rod of semi-infinite length at a point vertically above its right end (also true if point P lies vertically above the left end of semi-infinite rod), always points 45° away from the axis of the rod, regardless of how near or far the point is from the rod.” and
The value of equations (11) and (12) is half of the equation (8), which is obvious because if we join together the ends of two equally-charged semi-infinite rods, we get an infinite rod. The x components of the two rods are equal and opposite, so they cancel out and we get equation (8).

Case V :- To find Electric Field Strength when the point P lies vertically above a point S, which is at a distance d from the right end, on the axis of the finite charged rod

From the above figure…

$\displaystyle \sin {{\theta }_{R}}=\frac{-d}{\sqrt{{{R}^{2}}+{{d}^{2}}}}$

$\displaystyle \sin {{\theta }_{L}}=\frac{-(d+L)}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}}$

$\displaystyle \cos {{\theta }_{R}}=\frac{R}{\sqrt{{{R}^{2}}+{{d}^{2}}}}$ and

$\displaystyle \cos {{\theta }_{L}}=\frac{R}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}}$

Using the value of cosθ_L and cosθ_R, in equation (5), we get…

$\displaystyle {{E}_{x}}=\frac{k\lambda }{R}\left[ \frac{R}{\sqrt{{{R}^{2}}+{{d}^{2}}}}-\frac{R}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}} \right]$

⇒ $\displaystyle {{E}_{x}}=k\lambda \left[ \frac{1}{\sqrt{{{R}^{2}}+{{d}^{2}}}}-\frac{1}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}} \right]$ …..(13)

Similarly, using the value of sinθ_L and sinθ_R, in equation (6), we get…

$\displaystyle {{E}_{y}}=\frac{k\lambda }{R}\left[ \frac{-d}{\sqrt{{{R}^{2}}+{{d}^{2}}}}-\frac{-(d+L)}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}} \right]$

⇒ $\displaystyle {{E}_{y}}=\frac{k\lambda }{R}\left[ \frac{(d+L)}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}}-\frac{d}{\sqrt{{{R}^{2}}+{{d}^{2}}}} \right]$ …..(14)

Case VI :- To find Electric Field Strength when the point P lies on the axis of the finite charged rod at a distance d from the right end

From above figure as the point P lies on the axis of the rod so, R→0

Taking limit as R→0, sinθ_Rand sinθ_L becomes….

$\displaystyle \underset{R\to 0}{\mathop{\text{Lim}}}\,\left[ \sin {{\theta }_{R}}=\frac{-d}{\sqrt{{{R}^{2}}+{{d}^{2}}}} \right]\Rightarrow \sin {{\theta }_{R}}\to -1\text{ }$

$\displaystyle \underset{R\to 0}{\mathop{\text{Lim}}}\,\left[ \sin {{\theta }_{L}}=\frac{-(d+L)}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}} \right]\Rightarrow \sin {{\theta }_{L}}\to -1$

From equation (6), we get E_y = 0.

Similarly applying limit, R→0, and then using equation(13) we get…

$\displaystyle \underset{R\to 0}{\mathop{Lim}}\,\left[ {{E}_{x}} \right]=\underset{R\to 0}{\mathop{Lim}}\,\left[ \frac{k\lambda }{R}\left[ \frac{R}{\sqrt{{{R}^{2}}+{{d}^{2}}}}-\frac{R}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}} \right] \right]=\underset{R\to 0}{\mathop{Lim}}\,\left[ k\lambda \left[ \frac{1}{\sqrt{{{R}^{2}}+{{d}^{2}}}}-\frac{1}{\sqrt{{{R}^{2}}+{{(d+L)}^{2}}}} \right] \right]$

$\displaystyle \Rightarrow {{E}_{x}}=k\lambda \left[ \frac{1}{d}-\frac{1}{d+L} \right]=\frac{k\lambda L}{d\left( d+L \right)}$

$\displaystyle \Rightarrow {{E}_{x}}=\frac{kQ}{d\left( d+L \right)}$ …..(15)

[ Because λL = Q, total charge on the rod]

Note :- If d >>>>L, then, equation (15) becomes, $\displaystyle {{E}_{x}}=\frac{kQ}{{{d}^{2}}}$ , which is nothing but Field Intensity due to a point charge Q at a distance d. Because if d >>>>L, then the rod will behave as a point charge.

Electric Field Strength due to a Uniformly Charged Rod at a General Point

Electric Field Strength due to a Uniformly Charged Rod at a General Point

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