Important Derivations for Class 12 Physics Chapter 1 Electric Charges and Fields

Important Derivations for Class 12 Physics Chapter 1 Electric Charges and Fields

Q.1. Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

Solution:

Electric field intensity due to a thin infinite sheet of charge: Let σ be the surface density of charge and P be a point at a distance r from the sheet where has to be calculated. on either side is perpendicular to the sheet.

Imagine a cylinder of cross-sectional area ds around P and length 2r, piercing through the sheet. At the two edges, (or ).
At the curved surfaces $$\overrightarrow{\mathrm{E}} \perp \hat{\mathrm{n}}$$. So, there is no contribution to electric flux from the curved surfaces of the cylinder.
Electric flux over the edges $$=2 \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=2 \mathrm{EdS}$$
Total charge enclosed by the cylinder $$=\sigma \mathrm{dS}$$
By Gauss’s theorem, $$2EdS=\frac{\mathrm{q}}{\varepsilon_0}=\frac{\sigma \mathrm{dS}}{\varepsilon_0} \\\mathrm{E}=\frac{\sigma}{2 \varepsilon_0}$$
From the expression, it is clear that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

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Q.2. Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field.

Solution:

Let the dipole moment makes an angle with the direction of .
Force on the charge -q = -qE
Force on the charge q = +qE
These forces are equal in magnitude but act along parallel lines at different points hence form a couple which tends to rotate the dipole in the anticlockwise direction.
As, torque = moment of force.
$$\begin{aligned} & \tau= F \times A B \sin \theta=(\mathrm{qE})(2 \mathrm{a} \sin \theta)=(\mathrm{q} \times 2 \mathrm{a}) E \sin \theta \\ &(\mathrm{p}=\mathrm{q} \times 2 \mathrm{a}) \\ &=\mathrm{pE} \sin \theta \\ & \text { or, } \quad \vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}\end{aligned}$$

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Q.3. Derive an expression for the electric field due to an infinitely long straight wire of linear charge density λ C/m.

Solution:

Electric field due to an infinitely long straight wire:
Consider an infinitely long line charge having linear charge density λ. To determine its electric field at distance $r$, consider a cylindrical Gaussian surface of radius r and length ℓ coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface S1 and is directed radially outward.
Total flux through the cylindrical surface,
$$
\begin{aligned}
\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}} & =\oint_{\mathrm{S}_1} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}_1}+\oint_{\mathrm{S}_2} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}{ }_2+\oint_{\mathrm{S}_3} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}_3} \\
& =\oint_{\mathrm{S}_1} \mathrm{Eds}_1 \cdot \cos 0^{\circ}+\oint_{\mathrm{S}_2} \mathrm{Eds}_2 \cdot \cos 90^{\circ}+\oint_{\mathrm{S}_3} \mathrm{Eds}_3 \cdot \cos 90^{\circ} \\
& =\mathrm{E} \int \mathrm{ds}_1=\mathrm{E} \times 2 \pi \mathrm{r} l
\end{aligned}
$$
As λ is the charge per unit length and ℓ is the length of the wire,
so charge enclosed, $$\mathrm{q}=\lambda l$$
By Gauss’s theorem,
$$
\begin{array}{ll}
\oint_{\mathrm{S}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{\mathrm{q}}{\varepsilon_0} & \text { or, } \quad \mathrm{E} \times 2 \pi \mathrm{r} l=\frac{\lambda l}{\varepsilon_0} \\
\mathrm{E}=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}}
\end{array}
$$

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Q.4. Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.

Electric field intensity at point P due to +q (at B)
$$
\mathrm{E}_{\mathrm{B}}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{\mathrm{q}}{\left(\mathrm{r}^2+l^2\right)} \text { along BP }
$$
Electric field intensity at $\mathrm{P}$ due to $-\mathrm{q}$ charge (at $\mathrm{A}$ )
$$
\mathrm{E}_{\mathrm{A}}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{\mathrm{q}}{\left(\mathrm{r}^2+l^2\right)} \text { along PA }
$$
Clearly, in magnitude.
and can be resolved into two rectangular components.
Components of
(i) along PX
(ii) along PY
Components of $\mathrm{E}_{\mathrm{B}}$
(i) along PX
(ii) along YP
Vertical components being equal and opposite cancel each other. Therefore, net electric field intensity along PX
$$
\begin{aligned}
\mathrm{E} & =\mathrm{E}_{\mathrm{A}} \cos \theta+\mathrm{E}_{\mathrm{A}} \cos \theta \quad\left(\mathrm{E}_{\mathrm{A}}=\mathrm{E}_{\mathrm{B}}\right) \\
& =2 \mathrm{E}_{\mathrm{A}} \cos \theta \text { along PX } \\
& =2 \cdot \frac{1}{4 \pi \epsilon_0} \cdot \frac{\mathrm{q}}{\left(\mathrm{r}^2+\mathrm{l}^2\right)} \cdot \frac{1}{\sqrt{\mathrm{r}^2+\mathrm{l}^2}}
\end{aligned}
$$
or, $$E=\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{\left(r^2+l^2\right)^{3 / 2}} \, along\, PX$$
$$\quad( p=q \times 2 \mathrm{l})$$ If l << r so that it can be neglected, then $$ \begin{aligned} & E=\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^3} \text { along PX } \\ \quad & E \propto \frac{1}{r^3} \end{aligned} $$

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Q.5. A dipole, with a dipole moment of magnitude p, is in stable equilibrium in an electrostatic field of magnitude E. Find the work done in rotating this dipole to its position of unstable equilibrium.

For stable equilibrium, angle between p and E is 0⁰
For unstable equilibrium $$\theta_2=180^{\circ}$$.
Work done in rotating the dipole from angle $\latex theta_1$ to .
$$
W=p E\left(\cos \theta_1-\cos \theta_2\right)=p E\left(\cos 0^{\circ}-\cos 180^{\circ}\right)=2 p E
$$

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Q.6. Deduce the expression for the electric field E due to a system of two charge q1 and q2 with position vectors r1 and r2 at a point ‘r’ with respect to common origin.

Electric field intensity at point due to ,
$$
E_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{|A P|^3} A P
$$
Similarly, $$E_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{|B P|^3} B P$$
Therefore, Net electric field intensity at point P,
$$
\mathrm{E}=\mathrm{E}_1+\mathrm{E}_2=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{\left|r-r_1\right|^3}\left(r-r_1\right)+\frac{q_2}{\left|r-r_2\right|^3}\left(r-r_2\right)\right]
$$

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Q.7. Two charge –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x << d) perpendicular to the line joining the two fixed charged as shown in Fig. Show that q will perform simple harmonic oscillation and also find its time period.

To show that \( q \) will perform simple harmonic oscillation and derive the expression for its time period, we can follow these steps: Let the force on charge \( q \) due to one of the fixed charges (\(-q\)) be \( F_1 \) and the force due to the other fixed charge be \( F_2 \). The net force on charge \( q \) is given by: \[ F_{\text{net}} = F_1 + F_2 \] The force between two point charges is given by Coulomb’s Law: \[ F = \frac{k|q_1q_2|}{r^2} \] Where \( k \) is the Coulomb’s constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the separation between the charges. Let the distance of charge \( q \) from each of the fixed charges be \( d \). So, the force \( F_1 \) and \( F_2 \) can be written as: \[ F_1 = \frac{k|q(-q)|}{d^2} = -\frac{kq^2}{d^2} \] \[ F_2 = \frac{k|q(-q)|}{d^2} = -\frac{kq^2}{d^2} \] Since both forces are in opposite directions and have the same magnitude, the net force \( F_{\text{net}} \) on charge \( q \) will be zero. Now, when charge \( q \) is displaced slightly by \( x \) in the perpendicular direction (as shown in the figure), the net force \( F_{\text{net}} \) will not remain zero. There will be a restoring force that brings \( q \) back to the equilibrium position. The restoring force can be found using Hooke’s Law: \[ F_{\text{restoring}} = -kx \] where \( k \) is the effective force constant. The effective force constant \( k \) can be determined by considering the net force on charge \( q \) when it is displaced by \( x \). We can find this by using the concept of superposition: \[ F_{\text{net}} = F_1 + F_2 + F_{\text{restoring}} \] Since \( F_1 \) and \( F_2 \) were calculated earlier, we can substitute their values: \[ 0 = -\frac{kq^2}{d^2} -\frac{kq^2}{d^2} – kx \] Now, solving for \( k \): \[ k = \frac{2kq^2}{d^2} \] The angular frequency \( \omega \) of the simple harmonic oscillation can be written as: \[ \omega = \sqrt{\frac{k}{m}} \] Substitute the value of \( k \): \[ \omega = \sqrt{\frac{2kq^2}{md^2}} \] The time period \( T \) of the oscillation is given by: \[ T = \frac{2\pi}{\omega} \] Substitute the value of \( \omega \): \[ T = \frac{2\pi}{\sqrt{\frac{2kq^2}{md^2}}} \]

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Why students face difficulty in physics derivations?

There can be several reasons why students face difficulty in physics derivations, including:

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Remembering physics derivations can be a challenge, but here are some tips that may help:

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Remember, it takes time and consistent effort to remember physics derivations. Keep practicing and seeking help when needed, and you will likely see improvement over time.

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