Solution:

Ans. Consider an electric dipole of charges $+q$ and $-q$ separated by $2 x$ distance being placed in free space. Let $P$ be the point at which the electric field is to be determined due to the electric dipole.

Let the potential at point $P$ due to positive charge be $V_{+}$and the potential at point $P$ due to negative charge $V_{-}$

$$ \begin{aligned} & V_{+}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{(d-x)} \ & V_{-}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{-q}{(d+x)} \end{aligned} $$

$\therefore$ The total potential at point $P$ is given by

$$ \begin{aligned} V & =V_{+}+V_{-} \ V & =\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{d-x}-\frac{q}{d+x}\right) \ V & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 x q}{\left(d^{2}-x^{2}\right)} \ \therefore \quad 2 q x & =p \ V & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{\left(d^{2}-x^{2}\right)} \end{aligned} $$

Now, if $x«<d$

$$ \begin{aligned} x^{2} & \cong 0 \ V & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{d^{2}} \end{aligned} $$

AT Q. 5. Four point charges $Q, q, Q$ and $q$ are placed at the corners of a square of side ’ $a$ ’ as shown in the figure.

Find the

(i) resultant electric force on a charge $Q$, and

(ii) potential energy of this system.

[Delhi/O.D. CBSE 2018]

Ans. (i) Finding the resultant force on a charge $Q \quad 2$

(ii) Potential Energy of the system

(i) Let us find the force on the charge $Q$ at the point $C$ Force due to the other charge $Q$

$$ \begin{aligned} & F_{1}=\frac{1 Q Q^{2}}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{2}}{(a \sqrt{2})^{2}} \ & \overrightarrow{F_{1}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{Q^{2}}{2 a^{2}}\right)(\text { along } A C) \end{aligned} $$

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Force due to the charge $q$ (at $B$ ),

$$ \vec{F}{2}=\frac{1}{4 \pi \varepsilon{0}} \cdot \frac{q Q}{a^{2}} \text { along } B C $$

Force due to the charge $q$ (at $D$ ),

$$ \overrightarrow{F_{3}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q Q}{a^{2}} \text { along } D C $$

Resultant of these two equal forces $\vec{F}{2} & \overrightarrow{F{3}}$

$$ \left.\overrightarrow{F_{23}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q Q(\sqrt{2})}{a^{2}} \text { (along } A C\right) $$

$\therefore$ Net force on charge $Q$ ( at point $C$ )

$$ \vec{F}=\vec{F}{1}+\overrightarrow{F{23}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{a^{2}}\left[\frac{Q}{2}+\sqrt{2} q\right] $$

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This force is directed along $A C$.

( For the charge $Q$, at the point $A$, the force will have the same magnitude but will be directed along $C A$ )

[Note : Don’t deduct marks if the student does not write the direction of the net force , $F]$

(ii) Potential energy of the system

$$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon_{0}}\left[4 \frac{q Q}{a}+\frac{q^{2}}{a \sqrt{2}}+\frac{Q^{2}}{a \sqrt{2}}\right] \ & =\frac{1}{4 \pi \varepsilon_{0} a}\left[4 q Q+\frac{q^{2}}{\sqrt{2}}+\frac{Q^{2}}{\sqrt{2}}\right] \end{aligned} $$

[CBSE Marking Scheme 2018]