Solution:

Ans. (i) Derivation

(ii) Difference

(i) Consider origin at the centre of dipole. As per superposition principle, potential due to dipole will be the sum of potentials due to charges $q$ and $-q$

$$ V=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{r_{1}}-\frac{q}{r_{2}}\right] $$

Where,

$r_{1}$ and $r_{2}=$ distances of point $P$ from $q$ and $-q$.

From the arrangements,

$$ \begin{aligned} & \text { ments, } \ & r_{1}^{2}=r^{2}+a^{2}-2 \arccos \theta \ & r_{2}^{2}=r^{2}+a^{2}+2 \arccos \theta \end{aligned} \quad \begin{aligned} & 1 / 2 \ & 1 / 2 \end{aligned} $$

If $r$ is greater than $a$, and taking terms upto first order in $a / r$

Also,

$$ r_{2}^{2} \cong r^{2}\left[1+\frac{2 a \cos \theta}{r}\right] $$

With the help of Binomial theorem, keeping terms upto first order in $a / r$;

$$ \begin{aligned} \frac{1}{r_{1}} & \cong \frac{1}{r}\left[1-\frac{2 a \cos \theta}{r}\right]^{-\frac{1}{2}} \ & \cong \frac{1}{r}\left[1+\frac{a}{r} \cos \theta\right] \end{aligned} $$

$$ \begin{aligned} \frac{1}{r_{2}} & \cong \frac{1}{r}\left[1+\frac{2 a \cos \theta}{r}\right]^{-\frac{1}{2}} \ & \cong \frac{1}{r}\left[1-\frac{a}{r} \cos \theta\right] \end{aligned} $$

As

$$ \begin{aligned} p & =q a \ V & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q(2 a) \cos \theta}{r^{2}} \end{aligned} $$

$$ V=\frac{p}{4 \pi \varepsilon_{0}} \cdot \frac{\cos \theta}{r^{2}} $$

Now, $\quad p \cos \theta=\vec{p} \cdot \hat{r}$

where, $\hat{r}$ is unit vector along position vector. Hence electric potential of dipole for distances large compared to size of dipole is given as :

$$ V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\vec{p} \cdot \hat{r}}{r^{2}} \text { for } r»a $$

(ii) For potential at any point on axis, $\theta=[0, \pi]$

It is obsered that :

potentialis positive when $\theta=0$

poential is negative when $\theta=\pi$

Hence, electric potential falls at large distance, as

$\frac{1}{r^{2}}$ and not as $\frac{1}{r}$

[CBSE Marking Scheme 2017]

[AI Q. 2. (i) Obtain the expression for the potential to show due to a point charge.

(ii) Potential, due to an electric dipole (length $2 a$ ) varies as the ‘inverse square’ of the distance of the ‘field point’ from the centre of the dipole for $r>a$. U] [Comptt. Delhi. I, II, III 2016]

Ans. (i)

$1 / 2$

Consider a point charge ’ $Q$ ’ kept at point $O$. Let $P$ be the field point at distance $r$.

At some intermediate point $P^{\prime}$ the electrostatic force on the unit positive charge is :

$$ =\frac{Q \times 1}{4 \pi \varepsilon_{0} r^{\prime 2}} $$

Work done against this force from $r^{\prime}$ to $r^{\prime}+\Delta r^{\prime}$ is

$$ \Delta W=\frac{Q}{4 \pi \varepsilon_{0} r^{\prime 2}} \Delta r^{\prime} $$

$1 / 2$

Total work done ’ $W$ ’ by the external Force from $\infty$ to $r$

$$ W=-\int_{\infty}^{r} \frac{Q}{4 \pi \varepsilon_{0} r^{\prime 2}} \Delta r^{\prime} $$

$1 / 2$

$$ \begin{aligned} & W=\frac{Q}{4 \pi \varepsilon_{0} r} \ & V=W=\frac{Q}{4 \pi \varepsilon_{0} r} \ & V_{1}=\frac{-1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{(r+a)} \ & V_{2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{(r-a)} \ & V=V_{1}+V_{2}=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{-1}{(r+a)}+\frac{1}{(r-a)}\right] \ & =\frac{q \times 2 a}{4 \pi \varepsilon_{0}\left(r^{2}-a^{2}\right)} \end{aligned} $$

(AI Q. 3. (i) Two isolated metal spheres $A$ and $B$ have radii $R$ and $2 R$ respectively, and same charge $q$. Find which of the two spheres have greater :

(a) capacitance and

(b) energy density just outside the surface of the spheres.

(ii) (a) Show that the equipotential surfaces are closed together in the regions of strong field and far apart in the regions of weak field. Draw equipotential surfaces for an electric dipole.

(b) Concentric equipotential surfaces due to a charged body placed at the centre are shown. Identify the polarity of the charge and draw the electric field lines due to it.

U] [CBSE SQP 2015-16] Ans. (i) (a)

$$ C_{A}=4 \pi \varepsilon_{0} R, C_{B}=4 \pi \varepsilon_{0}(2 R) \quad 1 / 2 $$

$$ C_{B}>C_{A} $$

(b)

$$ U=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} $$

$1 / 2$

$$ E=\frac{\sigma}{\varepsilon_{0}}=\frac{\mathrm{Q}}{A \varepsilon_{0}} $$

$$ \begin{array}{ll} \therefore & U \propto \frac{1}{\mathrm{~A}^{2}} \ \therefore & U_{A}>U_{B} \end{array} $$

(ii) (a)

$$ E=-\frac{d V}{d r} $$

For same change in $d V, E \propto \frac{1}{d r}$

where, ’ $d r$ ’ represents the distance between equipotential surfaces.

Diagram of equipotential surface due to a dipole 1

$\Delta \mathrm{V}=0$

(b) Polarity of charge : - negative

The direction of electric field is radially inward.

[CBSE Marking Scheme 2015]