exemplar-phy-12-electrostatic-potential-and-capacitance

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<p><strong>Ques.</strong> A capacitor of $4 \mu F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 \Omega$. The amount of charge on the capacitor plates will be</p>
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<div class="fragment">
<p>(a) $0$</p>
<p>(b) $4 \mu C$</p>
<p>(c) $16 \mu C$</p>
<p>(d) $8 \mu C$</p>
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<p><strong>Thinking process :</strong></p>
<p>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.</p>
<p>Thus, potential difference across lower and middle branch of circuit is equal to the potential difference across capacitor of upper branch of circuit.</p>
<p><strong>Ans.</strong> (d)</p>
<p>Current flows through $2 \Omega$ resistance from left to right, is given by</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>$I=\frac{V}{R+r}=\frac{2.5 V}{2+0.5}=1 A$</p>
<p>The potential difference across $2 \Omega$ resistance $V=I R=1 \times 2=2 V$</p>
<p>Since, capacitor is in parallel with $2 \Omega$ resistance, so it also has $2 V$ potential difference across it.</p>
<p>The charge on capacitor</p>
<p>$q=C V=(2 \mu F) \times 2 V=8 \mu C$</p>
<p>Note The potential difference across $2 \Omega$ resistance solely occurs across capacitor as no potential drop occurs across $10 \Omega$ resistance.</p>
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<p><strong>Ques.</strong> A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge</p>
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<p>(a) remains a constant because the electric field is uniform</p>
<p>(b) increases because the charge moves along the electric field</p>
<p>(c) decreases because the charge moves along the electric field</p>
<p>(d) decreases because the charge moves opposite to the electric field</p>
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<p><strong>Thinking process :</strong></p>
<p>In this problem, the relationship between $E$ and $V$ is actualised.</p>
<p><strong>Ans.</strong> (c)</p>
<p>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.</p>
<p>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.</p>
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<p><strong>Ques.</strong> Figure shows some equipotential lines distributed in space. A charged object is moved from point $A$ to point $B$.</p>
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<p>(a) The work done in Fig. (i) is the greatest</p>
<p>(b) The work done in Fig. (ii) is least</p>
<p>(c) The work done is the same in Fig. (i), Fig. (ii) and Fig. (iii)</p>
<p>(d) The work done in Fig. (iii) is greater than Fig. (ii) but equal to that in</p>
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<p><strong>Ques.</strong> The electrostatic potential on the surface of a charged conducting sphere is $100 V$. Two statements are made in this regard $S_1$ at any point inside the sphere, electric intensity is zero. $S_2$ at any point inside the sphere, the electrostatic potential is $100 V$. Which of the following is a correct statement?</p>
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<p>(a) $S_1$ is true but $S_2$ is false</p>
<p>(b) Both $S_1$ and $S_2$ are false</p>
<p>(c) $S_1$ is true, $S_2$ is also true and $S_1$ is the cause of $S_2$</p>
<p>(d) $S_1$ is true, $S_2$ is also true but the statements are independant</p>
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<p><strong>Ans.</strong> (c)</p>
<p>In this problem, the electric field intensity $E$ and electric potential $V$ are related as</p>
<p>$E=-\frac{d V}{d r}$</p>
<p>Electric field intensity $E=0$ suggest that $\frac{d V}{d r}=0$</p>
<p>This imply that $V=$ constant.</p>
<p>Thus, $E=0$ inside the charged conducting sphere causes, the same electrostatic potential $100 V$ at any point inside the sphere.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong> (a)</p>
<p>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</p>
<p>$V=k_e \frac{q}{r}$</p>
<p>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.</p>
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<p><strong>Ques.</strong> A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_1$ and dielectric constant $K_1$ and the other has thickness $d_2$ and dielectric constant $K_2$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness $d(=d_1+d_2)$ and effective dielectric constant $K$. The $K$ is</p>
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<div class="fragment">
<p>(a) $\frac{K_1 d_1+K_2 d_2}{d_1+d_2} $</p>
<p>(b) $\frac{K_1 d_1+K_2 d_2}{K_1+K_2} $</p>
<p>(c) $\frac{K_1 K_2(d_1+d_2)}{(K_1 d_1+K_2 d_2)}  $</p>
<p>(d) $\frac{2 K_1 K_2}K_1+ $</p>
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<p><strong>Thinking process :</strong></p>
<p>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 $K_2$.</p>
<p><strong>Ans.</strong> (c)</p>
<p>The capacitance of parallel plate capacitor filled with dielectric block has thickness $d_1$ and dielectric constant $K_2$ is given by</p>
<p>$C_1=\frac{K_1 \varepsilon_0 A}{d_1}$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Similarly, capacitance of parallel plate capacitor filled with dielectric block has thickness $d_2$ and dielectric constant $K_2$ is given by</p>
<p>$C_2=\frac{K_2 \varepsilon_0 A}{d_2}$</p>
<p>Since, the two capacitors are in series combination, the equivalent capacitance is given by</p>
<p>$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}  $</p>
<p>$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} $</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong> (b, c, d)</p>
<img src="https://cdn.mathpix.com/snip/images/bzQodvyjv72nCCapRuViL1Cy3Pej9QxbudPJ2Qa0vwg.original.fullsize.png" height="155px">
<p>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.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>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 $z$.</p>
<p>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.</p>
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<p><strong>Thinking process :</strong></p>
<p>In this problem, we need a relation between the electric field intensity $E$ and electric potential $\vee$ given by</p>
<p>$E=-\frac{d V}{d r}$</p>
<p><strong>Ans.</strong> (a, b, c)</p>
<p>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.</p>
<p>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.</p>
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<p><strong>Ans.</strong> (c)</p>
<p>Work done in displacing a charge particle is given by $W    _{12}=q(V_2-V_1)$ 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 $W=0$.</p>
<p>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.</p>
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<p><strong>Ques.</strong> In the circuit shown in figure initially key $K_1$ is closed and key $K_2$ is open. Then $K_1$ is opened and $K_2$ is closed (order is important). [Take $Q_1^{\prime}$ and $Q_2^{\prime}$ as charges on $C_1$ and $C_2$ and $V_1$ and $V_2$ as voltage respectively.]</p>
<p>Then,</p>
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<div class="fragment">
<p>(a) charge on $C_1$ gets redistributed such that $V_1=V_2$</p>
<p>(b) charge on $C_1$ gets redistributed such that $Q_1^{\prime}=Q_2^{\prime}$</p>
<p>(c) charge on $C_1$ gets redistributed such that $C_1 V_1+C_2 V_2=C_1 E$</p>
<p>(d) charge on $C_1$ gets redistributed such that $Q_1^{\prime}+Q_2^{\prime}=Q$</p>
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<p><strong>Thinking process :</strong></p>
<p>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 $C_2$.</p>
<p><strong>Ans.</strong> (a, d)</p>
<p>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.,</p>
<p>$Q_1^{\prime}+Q_2^{\prime}=Q$</p>
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<p><strong>Ques.</strong> A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.</p>
<p>$A:$ Key $K$ is kept closed and plates of capacitors are moved apart using insulating handle.</p>
<p>$B:$ Key $K$ is opened and plates of capacitors are moved apart using insulating handle.</p>
<p>Choose the correct option(s).</p>
<div style='float: left; width: 50%'>
<div class="fragment">
<p>(a) In ${A} Q$ remains same but $C$ changes<br>
(b) In ${B} V$ remains same but $C$ changes<br>
(c) In ${A} V$ remains same and hence $Q$ changes<br>
(d) In ${B} Q$ remains same and hence $V$ changes</p>
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<p><strong>Thinking process :</strong></p>
<p>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.</p>
<p><strong>Ans.</strong> (c, d)</p>
<p>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).</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>The electric potentials on spheres due to their charge need to be written in terms of their charge densities.</p>
<p><strong>Ans.</strong></p>
<p>Since, the two spheres are at the same potential, therefore</p>
<p>$\frac{k q_1}{R_1}=\frac{k q_2}{R_2} \Rightarrow \frac{k q_1 R_1}{4 \pi R_1^{2}}=\frac{k q_2 R_2}{4 \pi R_2^{2}}$</p>
<p>or</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong> A test charge $q$ is made to move in the electric field of a point charge $Q$ along two different closed paths figure first path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?</p>
<img src="https://cdn.mathpix.com/snip/images/AIbvbv_1ma8wsWrGiq3GNvtf9dFyri86_z7c6Bf8-mc.original.fullsize.png" height="200px">
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<p><strong>Thinking process :</strong></p>
<p>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.</p>
<p><strong>Ans.</strong></p>
<p>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 $\frac{d V}{d r}$.</p>
<p>Consequently electric field comes into existence, which is given by as $E=-\frac{d V}{d r}$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>Here, the charge stored by the capacitor remains conserved after its disconnection from battery.</p>
<p><strong>Ans.</strong></p>
<p>The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant $K$ is given by</p>
<p>$C=\frac{K \varepsilon_0 A}{d}$, where signs are as usual.</p>
<p>The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum $K=1$.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>After disconnection from battery charge stored will remain the same due to conservation of charge.</p>
<p>The energy stored in an isolated charge capacitor $=\frac{q^{2}}{2 C}$; as $q$ is constant, energy stored $\propto$ $\frac{1}{C}$ and $C$ decreases with the removal of dielectric medium, therefore energy stored increases.</p>
<p>Since $q$ is constant and $V=\frac{q}{C}$ and $C$ decreases which in turn increases $V$ and therefore $E$ increases as $E=\frac{V}{d}$.</p>
<p>Note One of the very important questions with the competitive point of view.</p>
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<p><strong>Thinking process :</strong></p>
<p>The electric field $E=\frac{d V}{d r}$ suggest that electric potential decreases along the direction of electric field.</p>
<p><strong>Ans.</strong></p>
<p>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.</p>
<p>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.</p>
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<p><strong>Thinking process :</strong></p>
<p>The work done or PE stored in a system of charges can be obtained $U=W=q \times$ potential difference</p>
<p><strong>Ans.</strong></p>
<p>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</p>
<p>$V=k_e \int \frac{d q}{r}=k_e \int \frac{d q}{\sqrt{z^{2}+a^{2}}}$<br>
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</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong></p>
<p>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</p>
<p>$V=k_e \int \frac{d q}{r}=k_e \int \frac{d q}{\sqrt{x^{2}+a^{2}}}$</p>
<p>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</p>
<p>$V=\frac{k_e}{\sqrt{x^{2}+a^{2}}} \int d q=\frac{k_e Q}{\sqrt{x^{2}+a^{2}}}$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>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.</p>
<p>$V(r)-V(r_0)=-\int_r^{r} E . d l$</p>
<p><strong>Ans.</strong></p>
<p>Let the field lines must be radically outward. Draw a cylindrical Gaussian surface of radius $r$ and length $l$. Then, applying Gauss’ theorem</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Hence, if $r_0$ is the radius,</p>
<p>$V(r)-V(r_0)=-\int_{r_0}^{r} E . d l=\frac{\lambda}{2 \pi \varepsilon_0} \ln \frac{r_0}{r}$</p>
<p>Since,</p>
<p>$\int_0^{r} \frac{\lambda}{2 \pi \varepsilon_0 r} d r=\frac{\lambda}{2 \pi \varepsilon_0} \int_0^{r} \frac{1}{r} d r=\frac{\lambda}{2 \pi \varepsilon_0} \ln \frac{r}{r_0}$</p>
<p>For a given $V$,</p>
<p>$\ln \frac{r}{r_0}  =-\frac{2 \pi \varepsilon_0}{\lambda}[V(r)-V(r_0)] \newline $</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>In this problem, the dielectric of variable permittivity is used which gives new insight in the ordinary problem.</p>
<p><strong>Ans.</strong></p>
<p>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 $Q_1=C U$</p>
<p>Since, the capacitor with the dielectric has a capacitance $\varepsilon C$.</p>
<p>Hence, the charge on the capacitor is given by</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>The disc will be lifted when weight is balanced by electrostatic force.</p>
<p><strong>Ans.</strong></p>
<p>Assuming initially the disc is in touch with the bottom plate, so the entire plate is a equipotential.</p>
<p>The electric field on the disc, when potential difference $V$ is applied across it, given by</p>
<p>$E=\frac{V}{d}$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Let charge $q^{\prime}$ is transferred to the disc during the process,</p>
<p>Therefore by Gauss’ theorem,</p>
<p>$\therefore q^{\prime}  =-\varepsilon_0 \frac{V}{d} \pi^{2} $</p>
<p>Since, Gauss theorem states that</p>
<p>$\varphi  =\frac{q}{\varepsilon_0} \text{ or } q=\frac{\varepsilon_0}{\varphi} $</p>
<p>$ =\varepsilon E A=\frac{\varepsilon_0 V}{d} A $</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong> (a) In a quark model of elementary particles, a neutron is made of one up quarks [charge $(\frac{2}{3})$ e] and two down quarks [charges -$(\frac{2}{3})$ e]. Assume that they have a triangle configuration with side length of the order of 10$^{-15}$ m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.</p>
<p>(b) Repeat above exercise for a proton which is made of two up and one down quark.</p>
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<p><strong>Ans.</strong></p>
<p>This system is made up of three charges.</p>
<p>The potential energy of the system is equal to the algebraic sum of PE of each pair. So,</p>
<p>$U = \frac{1}{4 \pi e_0} \Big[\frac{q_uq_d}{r}- \frac{q_uq_d}{r}- \frac{q_uq_d}{r}\Big] $</p>
<p>$= \frac{9 \times 10^9}{10^{-15}} \big[{(1.6 \times 10^{-15})^2{1/3}^2 - (\frac{2}{3})(\frac{1}{3}) - (\frac{2}{3})(\frac{1}{3})}\big] $</p>
<p>$= 4.8 \times 0^{-13}\Big[{\frac{1}{9} -\frac{4}{9} }\Big]$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>Now applying law of conservation of charges is given by</p>
<p>$Q_1^{\prime}+Q_2^{\prime}  =Q_1+Q_2=5 Q_1 $</p>
<p>$ =5(\sigma .4 \pi R^{2}) $</p>
<p>After coming in contact, they acquire equal potentials. Therefore, we have</p>
<p>$\frac{1}{4 \pi \varepsilon_0} \frac{Q_1^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{Q_2^{\prime}}{R}$</p>
<p>On solving, we get</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong></p>
<p>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.</p>
<img src="https://cdn.mathpix.com/snip/images/05olQRbqzwLasnjjOoVzeoo5zakVM0lxx2bUs10M244.original.fullsize.png" height="150px">
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>The electric potential of each ring, of radius $r$ and width $d r$, have charge $d q$ is given by</p>
<p>$\sigma d A=\sigma 2 \pi r d r$</p>
<p>and potential is given by</p>
<p>$d V=\frac{k_e d q}{\sqrt{r^{2}+x^{2}}}=\frac{k_e \sigma 2 \pi r d r}{\sqrt{r^{2}+x^{2}}}$</p>
<p>where $k_e=\frac{1}{4 \pi \varepsilon_0}$ the total electric potential at $P$, is given by</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>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.</p>
<p><strong>Ans.</strong></p>
<p>Let any arbitrary point on the required plane is $(x, y, z)$. The two charges lies on $z$-axis at a separation of $2 d$.</p>
<p>The potential at the point $P$ due to two charges is given by</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p>$ \frac{q_1}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}+\frac{q_2}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}}=0 $</p>
<p>$\therefore  \frac{q_1}{\sqrt{x^{2}+y^{2}+(z-d)^{2}}}=\frac{-q_2}{\sqrt{x^{2}+y^{2}+(z+d)^{2}}} $</p>
<p>On squaring and simplifying, we get</p>
<p>$x^{2}+y^{2}+z^{2}+\frac{(q_1 / q_2)^{2}+1}{(q_1 / q_2)^{2}-1}(2 z d)+d^{2}-0$</p>
<p>This is the equation of a sphere with centre at</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong></p>
<p>Let third charge $+q$ is slightly displaced from mean position towards first charge. So, the total potential energy of the system is given by</p>
<p>$U  =\frac{1}{4 \pi \varepsilon_0} \frac{-q^{2}}{(d-x)}+\frac{-q^{2}}{(d+x)} $</p>
<p>$U  =\frac{-q^{2}}{4 \pi \varepsilon_0} \frac{2 d}{(d^{2}-x^{2})} $</p>
<p>$\frac{d U}{d x}  =\frac{-q^{2} . 2 d}{4 \pi \varepsilon_0} \cdot \frac{2 x}{(d^{2}-x^{2})^{2}} $</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><Success style="float:center; text-align:center;font-size:28px"><b>ELECTROSTATIC POTENTIAL AND CAPACITANCE</b> </Success></p>
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<p>$ =\frac{-2 d q^{2}}{4 \pi \varepsilon_0} \frac{1}{(d^{2}-x^{2})^{3}}[2(d^{2}-x^{2})^{2}-8 x^{2}] $</p>
<p>x = 0</p>
<p>$\frac{d^{2} U}{d x^{2}}  =\frac{-2 d q^{2}}{4 \pi \varepsilon_0} \quad \frac{1}{d^{6}}(2 d^{2}), \text{ which is }<0 $</p>
<p>This shows that system will be unstable equilibrium.</p>
<p>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 $\frac{d^{2} y}{d x^{2}}<0$</p>
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