Thinking Process

In this problem, the three parallel branches of circuit can be considered in parallel, combination with one-another. Therefore, potential difference across each branch is same. The capacitor offers infinite resistance in DC circuit, therefore no current flows through capacitor and $10 \Omega$ resistance, leaving zero potential difference across $10 \Omega$ resistance.

Thus, potential difference across lower and middle branch of circuit is equal to the potential difference across capacitor of upper branch of circuit.

Answer

(d) Current flows through $2 \Omega$ resistance from left to right, is given by

$$ I=\frac{V}{R+r}=\frac{2.5 \mathrm{~V}}{2+0.5}=1 \mathrm{~A} $$

The potential difference across $2 \Omega$ resistance $V=I R=1 \times 2=2 \mathrm{~V}$

Since, capacitor is in parallel with $2 \Omega$ resistance, so it also has $2 \mathrm{~V}$ potential difference across it.

The charge on capacitor

$$ q=C V=(2 \mu F) \times 2 V=8 \mu C $$

Note The potential difference across $2 \Omega$ resistance solely occurs across capacitor as no potential drop occurs across $10 \Omega$ resistance.

Thinking Process

In this problem, the relationship between $E$ and $V$ is actualised.

Answer

(c) The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.

The positively charged particle experiences electrostatic force along the direction of electric field i.e., from high electrostatic potential to low electrostatic potential. Thus, the work is done by the electric field on the positive charge, hence electrostatic potential energy of the positive charge decreases.

Answer

(c) The work done by a electrostatic force is given by $W_{12}=q\left(V_{2}-V_{1}\right)$. Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.

Answer

(c) In this problem, the electric field intensity $E$ and electric potential $V$ are related as

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

Electric field intensity $E=0$ suggest that $\frac{d V}{d r}=0$

This imply that $V=$ constant.

Thus, $E=0$ inside the charged conducting sphere causes, the same electrostatic potential $100 \mathrm{~V}$ at any point inside the sphere.

Note $V$ equals zero does not necessary imply that $E=0$ e.g., the electric potential at any point on the perpendicular bisector due to electric dipole is zero but $E$ not.

$E=0$ does not necessary imply that $V=0$ e.g., the electric field intensity at any point inside the charged spherical shell is zero but there may exist non-zero electric potential.

Answer

(a) In this problem, the collection of charges, whose total sum is not zero, with regard to great distance can be considered as a point charge. The equipotentials due to point charge are spherical in shape as electric potential due to point charge $q$ is given by

$$ V=k_{e} \frac{q}{r} $$

This suggest that electric potentials due to point charge is same for all equidistant points. The locus of these equidistant points, which are at same potential, form spherical surface.

Thinking Process

In this problem, the system can be considered as the series combination of two capacitors which are of thicknesses $d_{1}$ and filled with dielectric medium of dielectric constant $K_{1}$ and thicknesses $d_{2}$ and filled with dielectric medium of dielectric constant $K_{2}$.

Answer

(c) The capacitance of parallel plate capacitor filled with dielectric block has thickness $d_{1}$ and dielectric constant $K_{2}$ is given by

$$ C_{1}=\frac{K_{1} \varepsilon_{0} A}{d_{1}} $$

Similarly, capacitance of parallel plate capacitor filled with dielectric block has thickness $d_{2}$ and dielectric constant $K_{2}$ is given by

$$ C_{2}=\frac{K_{2} \varepsilon_{0} A}{d_{2}} $$

Since, the two capacitors are in series combination, the equivalent capacitance is given by

$$ \begin{gathered} \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \newline \newline\ C=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{\frac{K_{1} \varepsilon_{0} A}{d_{1}} \frac{K_{2} \varepsilon_{0} A}{d_{2}}}{\frac{K_{1} \varepsilon_{0} A}{d_{1}}+\frac{K_{2} \varepsilon_{0} A}{d_{2}}}=\frac{K_{1} K_{2} \varepsilon_{0} A}{K_{1} d_{2}+K_{2} d_{1}} \end{gathered} $$

But the equivalent capacitances is given by

$$ C=\frac{K \varepsilon_{0} A}{d_{1}+d_{2}} $$

On comparing, we have

$$ K=\frac{K_{1} K_{2}\left(d_{1}+d_{2}\right)}{K_{1} d_{2}+K_{2} d_{1}} $$

Note For the equivalent capacitance of the combination, thickness is equal to the separation between two plates i.e., $d_{1}+d_{2}$ and dielectric constant $K$.

Answer

$(b, c, d)$

Here, the figure electric field is always remain in the direction in which the potential decreases steepest. Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

The electric field in $z$-direction suggest that equipotential surfaces are in $x$-y plane. Therefore the potential is a constant for any $x$ for a given $z$, for any $y$ for a given $z$ and on the $x$-y plane for a given $z$.

Note The shape of equipotential surfaces depends on the nature and type of distribution of charge e.g., point charge leads to produce spherical surfaces whereas line charge distribution produces cylindrical equipotential surfaces.

Thinking Process

In this problem, we need a relation between the electric field intensity $E$ and electric potential $\vee$ given by

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

Answer

$(a, b, c)$

The electric field intensity $E$ is inversely proportional to the separation between equipotential surfaces. So, equipotential surfaces are closer in regions of large electric fields.

Since, the electric field intensities is large near sharp edges of charged conductor and near regions of large charge densities. Therefore, equipotential surfaces are closer at such places.

Answer

(c) Work done in displacing a charge particle is given by $W_{12}=q\left(V_{2}-V_{1}\right)$ and the line integral of electrical field from point 1 to 2 gives potential difference $V_{2}-V_{1}=-\int_{1}^{2} E$.dl For equipotential surface, $V_{2}-V_{1}=0$ and $W=0$.

Note If displaced charged particle is $+1 C$, then and only then option (b) is correct. But the NCERT exemplar book has given (b) as correct options which probably not so under given conditions.

Answer

$(b, c)$

The electric field intensity $E$ and electric potential $V$ are related as $E=0$ and for $V=$ constant, $\frac{d V}{d r}=0$

This imply that electric field intensity $E=0$.

Thinking Process

When key $K_{1}$ is closed and key $K_{2}$ is open, the capacitor $C_{1}$ is charged by cell and when $K$ is opened and $K_{2}$ is closed, the charge stored by capacitor $C_{1}$ gets redistributed between $C_{1}$ and $C_{2}$.

Answer

(a, $d)$

The charge stored by capacitor $C_{1}$ gets redistributed between $C_{1}$ and $C_{2}$ till their potentials become same i.e., $V_{2}=V_{1}$. By law of conservation of charge, the charge stored in capacitor $C_{1}$ when key $K_{1}$ is closed and key $K_{2}$ is open is equal to sum of charges on capacitors $C_{1}$ and $C_{2}$ when $K_{1}$ is opened and $K_{2}$ is closed i.e.,

$$ Q_{1}^{\prime}+Q_{2}^{\prime}=Q $$

Answer

( $a, b$ )

The charge resides on the outer surface of a closed charged conductor.

Thinking Process

The cell is responsible for maintaining potential difference equal to its emf across connected capacitor in every circumstance. However, charge stored by disconnected charged capacitor remains conserved.

Answer

$(c, d)$

Case A When key $K$ is kept closed and plates of capacitors are moved apart using insulating handle, the separation between two plates increases which in turn decreases its capacitance $C=\frac{K \varepsilon_{0} A}{d}$ and hence, the charge stored decreases as $Q=C V$ ( potential continue to be the same as capacitor is still connected with cell).

Case B When key $K$ is opened and plates of capacitors are moved apart using insulating handle, charge stored by disconnected charged capacitor remains conserved and with the decreases of capacitance, potential difference $V$ increases as $V=Q / C$.

Thinking Process

The electric potentials on spheres due to their charge need to be written in terms of their charge densities.

Answer

Since, the two spheres are at the same potential, therefore

$$ \frac{k q_{1}}{R_{1}}=\frac{k q_{2}}{R_{2}} \Rightarrow \frac{k q_{1} R_{1}}{4 \pi R_{1}^{2}}=\frac{k q_{2} R_{2}}{4 \pi R_{2}^{2}} $$

or

$$ \begin{gathered} \sigma_{1} R_{1}=\sigma_{2} R_{2} \Rightarrow \frac{\sigma_{1}}{\sigma_{2}}=\frac{R_{2}}{R_{1}} \newline \newline R_{2}>R_{1} \end{gathered} $$

This imply that $\sigma_{1}>\sigma_{2}$.

The charge density of the smaller sphere is more than that of the larger one.

Answer

The free electrons experiences electrostatic force in a direction opposite to the direction of electric field being is of negative charge. The electric field always directed from higher potential to lower travel.

Therefore, electrostatic force and hence direction of travel of electrons is from lower potential to region of higher potential.

Thinking Process

The capacity of conductor depend on its geometry i.e., length and breadth. For given charge potential $V \propto 1 / C$, so two adjacent conductors carrying the same charge of different dimensions may have different potentials.

Answer

Yes, if the sizes are different.

Answer

No, The absence of atmosphere around conductor prevents the phenomenon of electric discharge or potential leakage and hence, potential function do not have a maximum or minimum in free space.

Answer

As electric field is conservative, work done will be zero in both the cases.

Note Conservative forces (like electrostatic force or gravitational force) are those forces, work done by which depends only on initial position and final position of object viz charge, but not on the path through which it goes from initial position to final position.

Thinking Process

In this problem, we need to know that the electric field intensity $E$ and electric potential $V$ are related as $E=-\frac{d V}{d r}$ and the field lines are always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.

Answer

Let’s assume contradicting statement that the potential is not same inside the closed equipotential surface. Let the potential just inside the surface is different to that of the surface causing in a potential gradient $\frac{d V}{d r}$. Consequently electric field comes into existence, which is given by as $E=-\frac{d V}{d r}$.

Consequently field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside.

This contradict the original assumption. Hence, the entire volume inside must be equipotential.

Thinking Process

Here, the charge stored by the capacitor remains conserved after its disconnection from battery.

Answer

The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant $K$ is given by

$$ C=\frac{K \varepsilon_{0} A}{d}, \text { where signs are as usual. } $$

The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum $K=1$.

After disconnection from battery charge stored will remain the same due to conservation of charge.

The energy stored in an isolated charge capacitor $=\frac{q^{2}}{2 C}$; as $q$ is constant, energy stored $\propto$ $1 / C$ and $C$ decreases with the removal of dielectric medium, therefore energy stored increases. Since $q$ is constant and $V=q / C$ and $C$ decreases which in turn increases $V$ and therefore $E$ increases as $E=V / d$.

Note One of the very important questions with the competitive point of view.

Thinking Process

The electric field $E=\frac{d V}{d r}$ suggest that electric potential decreases along the direction of electric field.

Answer

Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.

Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Thinking Process

The work done or PE stored in a system of charges can be obtained $U=W=q \times$ potential difference

Answer

Let us take point $P$ to be at a distance $x$ from the centre of the ring, as shown in figure. The charge element $d q$ is at a distance $x$ from point $P$. Therefore, $V$ can be written as

$$ V=k_{e} \int \frac{d q}{r}=k_{e} \int \frac{d q}{\sqrt{z^{2}+a^{2}}} $$

where, $k=\frac{1}{4 \pi \varepsilon_{0}}$, since each element $d q$ is at the same distance from point $P$, so we have net potential

$$ V=\frac{k_{e}}{\sqrt{z^{2}+a^{2}}} \int d q=\frac{k_{e} Q}{\sqrt{z^{2}+a^{2}}} $$

Considering $-q$ charge at $P$, the potential energy is given by

$$ \begin{aligned} & U=W=q \times \text { potential difference } \ & U=\frac{k_{e} Q(-q)}{\sqrt{z^{2}+a^{2}}} \end{aligned} $$

or

$$ \begin{aligned} U & =\frac{1}{4 \pi \varepsilon_{0}} \frac{-Q q}{\sqrt{z^{2}+a^{2}}} \ & =\frac{1}{4 \pi \varepsilon_{0} a} \frac{-Q q}{\sqrt{1+\frac{z^{2}}{a}}} \end{aligned} $$

This is the required expression.

The variation of potential energy with $z$ is shown in the figure. The charge $-q$ displaced would perform oscillations.

Nothing can be concluded just by looking at the graph.

Answer

Let us take point $P$ to be at a distance $x$ from the centre of the ring,as shown in figure. The charge element $d q$ is at a distance $x$ from point $P$. Therefore, $V$ can be written as

$$ V=k_{e} \int \frac{d q}{r}=k_{e} \int \frac{d q}{\sqrt{x^{2}+a^{2}}} $$

where, $k_{e}=\frac{1}{4 \pi \varepsilon_{0}}$, since each element $d q$ is at the same distance from point $P$, so we have net potential

$$ V=\frac{k_{e}}{\sqrt{x^{2}+a^{2}}} \int d q=\frac{k_{e} Q}{\sqrt{x^{2}+a^{2}}} $$

The net electric potential

$$ V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{\sqrt{x^{2}+a^{2}}} $$

Thinking Process

The electric field due to line charge need to be obtained in order to find the potential at distance $r$ from the line charge. As line integral of electric field gives potential difference between two points.

$$ V(r)-V\left(r_{0}\right)=-\int_{r}^{r} E . d l $$

Answer

Let the field lines must be radically outward. Draw a cylindrical Gaussian surface of radius $r$ and length $l$. Then, applying Gauss’ theorem

or

$$ \int \mathrm{E} . \mathrm{dS}=\frac{1}{\varepsilon_{0}} \lambda l $$

$$ E_{r} 2 \pi r l=\frac{1}{\varepsilon_{0}} \lambda l \Rightarrow E_{r}=\frac{\lambda}{2 \pi \varepsilon_{0} r} $$

Hence, if $r_{0}$ is the radius,

$$ V(r)-V\left(r_{0}\right)=-\int_{r_{0}}^{r} E \cdot d l=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \frac{r_{0}}{r} $$

Since,

$$ \int_{0}^{r} \frac{\lambda}{2 \pi \varepsilon_{0} r} d r=\frac{\lambda}{2 \pi \varepsilon_{0}} \int_{0}^{r} \frac{1}{r} d r=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \frac{r}{r_{0}} $$

For a given $V$,

$$ \begin{aligned} \ln \frac{r}{r_{0}} & =-\frac{2 \pi \varepsilon_{0}}{\lambda}\left[V(r)-V\left(r_{0}\right)\right] \newline \ r & =r_{0} e^{-2 \pi \varepsilon_{0} V r_{0} / \lambda} e+2 \pi \varepsilon_{0} V(r) / \lambda \newline \ r & =r_{0} e^{-2 \pi \varepsilon_{0}\left[V_{(r)}-V_{\left(r_{0}\right)}\right] \lambda} \end{aligned} $$

The equipotential surfaces are cylinders of radius.

Thinking Process

The net electric potential at any point due to system of point charges is equal to the algebraic sum of electric potential due to each individual charges.

Answer

Let the required plane lies at a distance $x$ from the origin as shown in figure.

The potential at the point $P$ due to charges is given by

$$ \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{\left[(x+d / 2)^{2}+h^{2}\right]^{1 / 2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{\left[(x-d / 2)^{2}+h^{2}\right]^{1 / 2}} $$

If net electric potential is zero, then

Or

$$ \frac{1}{\left[(x+d / 2)^{2}+h^{2}\right]^{1 / 2}}=\frac{1}{\left[(x-d / 2)+h^{2}\right]^{1 / 2}} $$

$$ \begin{aligned} \Rightarrow & x^{2}-d x+d^{2} / 4 & =x^{2}+d x+d^{2} / 4 \ \text { Or } & 2 d x & =0 \Rightarrow x=0 \end{aligned} $$

The equation of the required plane is $x=0$ i.e., $y$-z plane.

Thinking Process

In this problem, the dielectric of variable permittivity is used which gives new insight in the ordinary problem.

Answer

Assuming the required final voltage be $U$. If $C$ is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is given by $Q_{1}=C U$

Since, the capacitor with the dielectric has a capacitance $\varepsilon C$. Hence, the charge on the capacitor is given by

$$ Q_{2}=\varepsilon C U=(\alpha U) C U=\alpha C U $$

The initial charge on the capacitor is given by

$$ \begin{array}{rlrl} \text { From the conservation of charges,\newline } \newline Q_{0} =C U_{0} \newline \ \text { Or } =Q_{1}+Q_{2} \newline \ C C U_{0} =C U+\alpha C U^{2} \newline \ \Rightarrow \alpha U^{2}+U-U_{0} =0 \newline \ \therefore U =\frac{-1 \pm \sqrt{1+4 \alpha U_{0}}}{2 \alpha} \end{array} $$

On solving for $U_{0}=78 \mathrm{~V}$ and $a=2 / \mathrm{V}$, we get

$$ U=6 \mathrm{~V} $$

Thinking Process

The disc will be lifted when weight is balanced by electrostatic force.

Answer

Assuming initially the disc is in touch with the bottom plate, so the entire plate is a equipotential.

The electric field on the disc, when potential difference $V$ is applied across it, given by

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

Let charge $q^{\prime}$ is transferred to the disc during the process,

Therefore by Gauss’ theorem,

$$ \therefore \quad \begin{aligned} q^{\prime} & =-\varepsilon_{0} \frac{V}{d} \pi^{2} \\ \text { Since, Gauss theorem states that } \quad \varphi & =\frac{q}{\varepsilon_{0}} \text { or } q=\frac{\varepsilon_{0}}{\varphi} \\ & =\varepsilon E A=\frac{\varepsilon_{0} V}{d} A \end{aligned} $$

The force acting on the disc is

$$ -\frac{V}{d} \times q^{\prime}=\varepsilon_{0} \frac{V^{2}}{d^{2}} \pi r^{2} $$

If the disc is to be lifted, then

$$ \varepsilon_{0} \frac{V^{2}}{d^{2}} \pi r^{2}=m g \Rightarrow V=\sqrt{\frac{m g d^{2}}{\pi \varepsilon_{0} r^{2}}} $$

This is the required expression.

Answer

This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of PE of each pair. So,

$$U = \frac{1}{4 \pi e_0} \Big[\frac{q_uq_d}{r}- \frac{q_uq_d}{r}- \frac{q_uq_d}{r}\Big] \newline \newline$$ $$= \frac{9 \times 10^9}{10^{-15}} [{(1.6 \times 10^{-15})^2{1/3}^2 - (2/3)(1/3) -(2/3)(1/3)}] \newline $$ $$= 4.8 \times 0^{-13}\Big[{\frac{1}{9} -\frac{4}{9} }\Big]$$ $$ = -7.68 \times 10^{-14} J$$ $$ = 4.8 \times 10^5 eV = 0.48meV \newline = 5.11 \times 10^{-4} (m_nc^2)$$

Answer

The charges on two metal spheres, before coming in contact, are given by

$$ \begin{aligned} Q & =\sigma .4 \pi R^{2} \ Q_{2} & =\sigma .4 \pi\left(2 R^{2}\right) \ & =4\left(\sigma .4 \pi R^{2}\right)=4 Q_{1} \end{aligned} $$

Let the charges on two metal spheres, after coming in contact becomes $Q_{1}^{\prime}$ and $Q_{2}^{\prime}$.

Now applying law of conservation of charges is given by

$$ \begin{aligned} Q_{1}^{\prime}+Q_{2}^{\prime} & =Q_{1}+Q_{2}=5 Q_{1} \ & =5\left(\sigma .4 \pi R^{2}\right) \end{aligned} $$

After coming in contact, they acquire equal potentials. Therefore, we have

$$ \frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{2}^{\prime}}{R} $$

On solving, we get

$$ \begin{array}{ll} \therefore & Q_{1}^{\prime}=\frac{5}{3}\left(\sigma \cdot 4 \pi R^{2}\right) \text { and } Q_{2}^{\prime}=\frac{10}{3}\left(\sigma \cdot 4 \pi R^{2}\right) \ \therefore & \sigma_{1}=5 / 3 \sigma \text { and } \ \therefore & \sigma_{2}=\frac{5}{6} \sigma \end{array} $$

Answer

In the circuit, when initially $K_{1}$ is closed and $K_{2}$ is open, the capacitors $C_{1}$ and $C_{2}$ acquires potential difference $V_{1}$ and $V_{2}$ respectively. So, we have

and

$$ \begin{aligned} V_{1}+V_{2} & =E \ V_{1}+V_{2} & =9 V \ V & \propto 1 / C \ V_{1}: V_{2} & =1 / 6: 1 / 3 \end{aligned} $$

Also, in series combination ,

On solving

$$ \begin{array}{ll} \Rightarrow &V_{1}=3 V \text { and } V_{2}=6 V \newline \ \therefore &Q_{1}=C_{1} V_{1}=6 C \times 3=18 \mu C \newline\ &Q_{2}=9 \mu C \text { and } Q_{3}=0 \newline \end{array} $$

Then, $K_{1}$ was opened and $K_{2}$ was closed, the parallel combination of $C_{2}$ and $C_{3}$ is in series with $C_{1}$.

$$ Q_{2}=Q_{2}^{\prime}+Q_{3} $$

and considering common potential of parallel combination as $V$, then we have

$$ \begin{aligned} C_{2} V+C_{3} V & =Q_{2} \newline\ V & =\frac{Q_{2}}{C_{2}+C_{3}}=(3 / 2) V \newline \ Q_{2}^{\prime} & =(9 / 2) \mu C \newline\ Q_{3} & =(9 / 2) \mu C \newline \end{aligned} $$

Answer

Let the point $P$ lies at a distance $x$ from the centre of the disk and take the plane of the disk to be perpendicular to the $\mathbf{x}$-axis. Let the disc is divided into a number of charged rings as shown in figure.

The electric potential of each ring, of radius $r$ and width $d r$, have charge $d q$ is given by

$$ \sigma d A=\sigma 2 \pi r d r $$

and potential is given by

(Refer the solution of 23.)

$$ d V=\frac{k_{e} d q}{\sqrt{r^{2}+x^{2}}}=\frac{k_{e} \sigma 2 \pi r d r}{\sqrt{r^{2}+x^{2}}} $$

where $k_{e}=\frac{1}{4 \pi \varepsilon_{0}}$ the total electric potential at $P$, is given by

$$ \begin{aligned} V & =\pi k_{e} \sigma \int_{0}^{a} \frac{2 r d r}{\sqrt{r^{2}+x^{2}}}=\pi k_{e} \sigma \int_{0}^{a}\left(r^{2}+x^{2}\right)^{-1 / 2} 2 r d r \newline \ V & =2 \pi k_{e} \sigma\left[\left(x^{2}+a^{2}\right)^{1 / 2}-x\right] \newline \ k_{e} & =\frac{1}{4 \pi \varepsilon_{0}} \newline \ V & =\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Q}{a^{2}}\left[\sqrt{x^{2}+a^{2}}-x\right] \end{aligned} $$

So, we have by substring

Note You may take $a=$ Rin this problem.

Thinking Process

Here, 3-dimensional imagination is required to actualise the problem. Also, the net electric potential at any point due to system of point charges is equal to the algebraic sum of electric potential due to each individual charges.

Answer

Let any arbitrary point on the required plane is $(x, y, z)$. The two charges lies on $z$-axis at a separation of $2 d$.

The potential at the point $P$ due to two charges is given by

$$ \begin{array}{ll} & \frac{q_{1}}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}+\frac{q_{2}}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}}=0 \newline \ \therefore & \frac{q_{1}}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}=\frac{-q_{2}}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}} \end{array} $$

On squaring and simplifying, we get

$$ x^{2}+y^{2}+z^{2}+\frac{\left(q_{1} / q_{2}\right)^{2}+1}{\left(q_{1} / q_{2}\right)^{2}-1}(2 z d)+d^{2}-0 $$

This is the equation of a sphere with centre at

$$ 0,0,-2 d \frac{q_{1}^{2}+q_{2}^{2}}{q_{1}^{2}-q_{2}^{2}} $$

Note The centre and radius of sphere $(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=r^{2}$ is $(a, b, c)$ and $r$ respectively.

Answer

Let third charge $+q$ is slightly displaced from mean position towards first charge. So, the total potential energy of the system is given by

$$ \begin{aligned} U & =\frac{1}{4 \pi \varepsilon_{0}} \frac{-q^{2}}{(d-x)}+\frac{-q^{2}}{(d+x)} \newline \ U & =\frac{-q^{2}}{4 \pi \varepsilon_{0}} \frac{2 d}{\left(d^{2}-x^{2}\right)} \newline \ \frac{d U}{d x} & =\frac{-q^{2} \cdot 2 d}{4 \pi \varepsilon_{0}} \cdot \frac{2 x}{\left(d^{2}-x^{2}\right)^{2}} \end{aligned} $$

The system will be in equilibrium, if

$$ F=-\frac{d U}{d x}=0 $$

On solving, $x=0$. So for, $+q$ charge to be in stable/unstable equilibrium, finding second derivative of PE.

At

$$ \begin{aligned} \frac{d^{2} U}{d x^{2}} & =\frac{-2 d q^{2}}{4 \pi \varepsilon_{0}} \quad \frac{2}{\left(d^{2}-x^{2}\right)^{2}}-\frac{8 x^{2}}{\left(d^{2}-x^{2}\right)^{3}} \newline \ & =\frac{-2 d q^{2}}{4 \pi \varepsilon_{0}} \frac{1}{\left(d^{2}-x^{2}\right)^{3}}\left[2\left(d^{2}-x^{2}\right)^{2}-8 x^{2}\right] \newline \ x & =0 \newline \ \frac{d^{2} U}{d x^{2}} & =\frac{-2 d q^{2}}{4 \pi \varepsilon_{0}} \quad \frac{1}{d^{6}}\left(2 d^{2}\right), \text { which is }<0 \end{aligned} $$

This shows that system will be unstable equilibrium.

Note For function $y=f(x)$, on solving $\frac{d y}{d x}=0$ gives critical points i.e., points of local maxima or local minima. If for any critical point, this imply that $y$ acquires maximum value at $x=x_{1}, x=x_{1}$ $\frac{d^{2} y}{d x^{2}}>0$ this imply that $y$ acquires minimum value at $x=x_{1}$ and for $\frac{d^{2} y}{d x^{2}}<0$