Derive coulomb's law from gauss's law

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According to gauss's law, the total electric flux (ϕ) through any close surface in free space is equal to 1/∈₀ times the total electric charge (q) enclosed by the surface.

Derive coulombs law from gausss law que phy

Mathematically,

\(\phi = \oint _S \overrightarrow{E} . \overrightarrow{ds} = \frac{q}{ \in _0}\)

According to coulomb’s law, the magnitude of the force of attraction and repulsion between any two point charges at rest is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

Mathematically, it is given by:

\(F=k \frac{q_1q_2}{r_2}\)

Derivation of coulomb’s law from gauss’s law

Consider two point charges q₁ and q₂ are separated by distance r in a vacuum.

Derivation of coulomb’s law from gauss’s law

Let E be the magnitude of the electric field at distance r from charge q₁. Therefore, the force(F) experienced by charge q₂ due to electric field(E) is given by:

F = q₂E …(i)

Draw a gaussian surface of radius r taking the center as the location of charge q₁.

The intensity of the electric field is constant at every point of this surface.

According to gauss’s law, the net electric flux through this surface is given by

\(\oint _S \overrightarrow{E} . \overrightarrow{ds} = \frac{q_1}{ \in _0}\)

⇒ \(\oint _S\) Eds cosθ = \(\frac{q_1}{ \in _0}\)

Direction of the electric field is always perpendicular to the surface. So, the angle between E and ds is 0.

⇒ \(\oint _S\)Eds cos0 = \(\frac{q_1}{ \in _0}\)

⇒ \(\oint _S\) Eds = \(\frac{q_1}{ \in _0}\)

Since the electric field is constant at every point of the gaussian surface, therefore we can write

E \(\oint _S\) ds = \(\frac{q_1}{ \in _0}\)

But, \(\oint _S\) ds = 4πr² is the surface area of sphere

⇒ E x 4πr²= \(\frac{q_1}{ \in _0}\)

Using equation (i), we get

\(\frac{F}{q_2}\) x 4πr² = \(\frac{q_1}{ \in _0}\)

⇒ F = \(\frac{1}{4 \pi \in _o} \frac{q_1q_2}{r_2}\)

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Derive coulomb's law from gauss's law

CBSE CLASS XII Related Questions

1.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5.
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Derive coulomb's law from gauss's law

CBSE CLASS XII Related Questions

1.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5. Write any two features of nuclear forces.

6. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

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Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

3.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

5.
Write any two features of nuclear forces.

6.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.