Why is electric flux through ring same as that of electric flux through disc?

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The electric flux through a ring is the same as the electric flux through a disc, provided they both have the same radius and the same total charge enclosed within them. This is known as Gauss's law, which states that the electric flux through a closed surface is proportional to the charge enclosed within the surface.

To understand why this is the case, consider a ring and a disc of the same radius, centered on the same axis. Let us assume that both the ring and the disc have a total charge Q enclosed within them.

Ring and a Disc of Same Radius

Ring and a Disc of Same Radius

The electric flux through the ring can be found by considering a small element of the ring, which is essentially a strip of infinitesimal width.
The electric field at any point on this strip is perpendicular to the strip, and the magnitude of the electric field is given by Coulomb's law.
The total flux through the ring is then found by integrating the electric field over the entire ring.
Similarly, the electric flux through the disc can be found by considering a small element of the disc, which is essentially a circular patch of infinitesimal area.
The electric field at any point on this patch is perpendicular to the patch, and the magnitude of the electric field is given by Coulomb's law.
The total flux through the disc is then found by integrating the electric field over the entire disc.

Now, since the ring and the disc have the same radius and the same total charge enclosed within them, the magnitude of the electric field at any point on the ring or the disc will be the same. Therefore, the integral of the electric field over the entire ring will be the same as the integral of the electric field over the entire disc, and hence the electric flux through the ring and the disc will be the same.

Therefore, the electric flux through a ring is the same as the electric flux through a disc, provided they both have the same radius and the same total charge enclosed within them. This is because the magnitude of the electric field at any point on the ring or the disc will be the same, and the integral of the electric field over the entire ring or disc will also be the same.

Also check:

Important Concepts Related to Electric Flux
Continuous Charge Distribution	Unit of Electric Field	Physical Significance of Electric Field
C harge density formula	Parallel Combination of Resistances	Electric Charges and Fields

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Why is electric flux through ring same as that of electric flux through disc?

CBSE CLASS XII Related Questions

2.
Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

4.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

5.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

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Why is electric flux through ring same as that of electric flux through disc?

CBSE CLASS XII Related Questions

2.Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

3.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

4.The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

5.A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

4.
The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.

5.
A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.