Electric Field and Voltage due to a uniformly charged Arc

Electric Field Due to a Uniformly Charged Arc

Electric Field Due to a Uniformly Charged Arc

Consider a uniformly charged arc of radius \(R\), total charge \(Q\), and angle subtended \(\alpha\) at the center. We derive the electric field at the center of the arc.

Electric Field

Only the horizontal components survive due to symmetry.

\[dE_x = dE\cos\theta\] \[dE_x=\frac{k\,dq}{R^2}\cos\theta\] \[dE_x=\frac{k}{R^2}\,\frac{Q}{R.\alpha}.dl\cos\theta\] \[dE_x=\frac{k}{R^2}\,\frac{Q}{R.\alpha}. R d\theta\cos\theta\] \[E=\frac{k}{R^2}\,\frac{Q}{\alpha} \int_{-\alpha/2}^{+\alpha/2}\cos\theta\, d\theta\] \[E=\frac{k}{R^2}\,\frac{Q}{\alpha} \left[\sin\theta\right]_{-\alpha/2}^{+\alpha/2}\] \[E=\frac{k}{R^2}\,\frac{Q}{\alpha} \left(\sin\frac{\alpha}{2}+\sin\frac{\alpha}{2}\right)\] \[\boxed{E=\frac{2kQ}{R^2\alpha} \sin\frac{\alpha}{2}}\]

In terms of $\lambda$ Linear Charge Density

\[\lambda = \frac{Q}{R\,\alpha}\] \[E=\frac{2k\lambda}{R}\sin\frac{\alpha}{2}\]

Electric Potential

All charge elements are at the same distance \(R\)from the center.

\[dV=\frac{k\,dq}{R}\] \[V=\int \frac{k\,dq}{R}\] \[V=\frac{k}{R}\int dq\] \[\boxed{V=\frac{kQ}{R}}\]

Special Cases

Quarter Circle, $\alpha = \frac{\pi}{2}$ \[E=\frac{2\sqrt{2}kQ}{\pi R^2}\] Semicircle, $\alpha = \pi$ \[E=\frac{2kQ}{\pi R^2}\] Full Circle $\alpha = 2\pi$ \[E = 0\] \[V = \frac{kQ}{R}\]

Detailed Explanation

Linear Charge Density

Total length of the arc:

\[ L = R\alpha \]

Therefore linear charge density:

\[ \lambda = \frac{Q}{R\alpha} \]

Small Charge Element

Take a small charge element \(dq\) on the arc.

\[ dE = \frac{k\,dq}{R^2} \]

Only the horizontal components survive due to symmetry.

\[ dE_x = dE\cos\theta \]

Substituting:

\[ dE_x = \frac{k\,dq}{R^2}\cos\theta \]

Substituting Charge Element

For a small angular width \(d\theta\):

\[ dq = \lambda R\, d\theta \]

Substitute into electric field expression:

\[ dE_x = \frac{k}{R^2} (\lambda R\,d\theta)\cos\theta \]

\[ dE_x = \frac{k\lambda}{R} \cos\theta\, d\theta \]

Integration

Integrate from:

\[ -\frac{\alpha}{2} \quad \text{to} \quad +\frac{\alpha}{2} \]

\[ E = \frac{k\lambda}{R} \int_{-\alpha/2}^{+\alpha/2} \cos\theta\, d\theta \]

\[ E = \frac{k\lambda}{R} \left[ \sin\theta \right]_{-\alpha/2}^{+\alpha/2} \]

\[ E = \frac{k\lambda}{R} \left( \sin\frac{\alpha}{2} + \sin\frac{\alpha}{2} \right) \]

\[ E = \frac{2k\lambda}{R} \sin\frac{\alpha}{2} \]

Final Expression for Electric Field

Using:

\[ \lambda = \frac{Q}{R\alpha} \]

Therefore:

\[ \boxed{ E = \frac{2kQ}{R^2\alpha} \sin\frac{\alpha}{2} } \]

Electric Potential Due to Charged Arc

All charge elements are at the same distance \(R\) from the center.

\[ dV = \frac{k\,dq}{R} \]

Integrating:

\[ V = \int \frac{k\,dq}{R} \]

\[ V = \frac{k}{R} \int dq \]

\[ V = \frac{kQ}{R} \]

Final Expression for Potential

\[ \boxed{ V = \frac{kQ}{R} } \]

Special Cases

Semicircle

\[ \alpha = \pi \]

\[ E = \frac{2kQ}{\pi R^2} \]

Full Circle

\[ \alpha = 2\pi \]

\[ E = 0 \]

\[ V = \frac{kQ}{R} \]

Quarter Circle

\[ \alpha = \frac{\pi}{2} \]

\[ E = \frac{2\sqrt{2}kQ}{\pi R^2} \]

Important Notes for JEE and NEET

• Electric field is a vector quantity.
• Potential is a scalar quantity.
• For a complete ring, electric field at the center becomes zero.
• Potential at the center of a ring is non-zero.
• Symmetry plays a crucial role in electric field derivations.