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

CBSE CLASS XII Related Questions

1.
In the circuit, three ideal cells of e.m.f. \( V \), \( V \), and \( 2V \) are connected to a resistor of resistance \( R \), a capacitor of capacitance \( C \), and another resistor of resistance \( 2R \) as shown in the figure. In the steady state, find (i) the potential difference between P and Q, (ii) the potential difference across capacitor C.

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2.
Briefly explain how and where the displacement current exists during the charging of a capacitor.

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Two coherent light waves, each of intensity \( I_0 \), superpose and produce an interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is \( \phi \). Write its maximum and minimum possible values.

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4.
Two point charges Q and \( -q \) are held \( r \) distance apart in free space. A uniform electric field \( \vec{E} \) is applied in the region perpendicular to the line joining the two charges. Which one of the following angles will the direction of the net force acting on charge \( -q \) make with the line joining Q and \( -q \)?

\( \tan^{-1} \left( \frac{4\pi \epsilon_0 E r^2}{Q} \right) \)
\( \cot^{-1} \left( \frac{4\pi \epsilon_0 E r^2}{Q} \right) \)
\( \tan^{-1} \left( \frac{QE}{4\pi \epsilon_0 r^2} \right) \)
\( \cot^{-1} \left( \frac{QE}{4\pi \epsilon_0 r^2} \right) \)

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5.
In a Young’s double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

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6.
Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.

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

CBSE CLASS XII Related Questions

2.
Briefly explain how and where the displacement current exists during the charging of a capacitor.

3.
Two coherent light waves, each of intensity \( I_0 \), superpose and produce an interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is \( \phi \). Write its maximum and minimum possible values.

5.
In a Young’s double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

6.
Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.

Similar Physics Concepts

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

CBSE CLASS XII Related Questions

2.Briefly explain how and where the displacement current exists during the charging of a capacitor.

3.Two coherent light waves, each of intensity \( I_0 \), superpose and produce an interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is \( \phi \). Write its maximum and minimum possible values.

5.In a Young’s double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

6.Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Briefly explain how and where the displacement current exists during the charging of a capacitor.

3.
Two coherent light waves, each of intensity \( I_0 \), superpose and produce an interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is \( \phi \). Write its maximum and minimum possible values.

5.
In a Young’s double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

6.
Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits.