A solid conducting sphere having a charge $Q$ is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$. If the shell is now given a charge of $-3Q$, the new potential difference between the same two surfaces is......$V$
IIT 1989, Diffcult
Download our app for free and get started
(a) In case of a charged conducting sphere
${V_{{\rm{inside}}}} = {V_{{\rm{centre }}}} = {V_{{\rm{surface}}}} = \frac{1}{{4\pi {\varepsilon _o}}}.\frac{q}{R}$, ${V_{{\rm{outside}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{r}$
If $a$ and $b$ are the radii of sphere and spherical shell respectively, then potential at their surface will be
${V_{{\rm{sphere }}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{a}$ and ${V_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{b}$
$\therefore $$V = {V_{{\rm{sphere }}}} - {V_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{Q}{a} - \frac{Q}{b}} \right]$
Now when the shell is given charge $(-3Q)$, then the potential will be
$V{'_{{\rm{sphere}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} + \frac{{( - 3Q)}}{b}} \right],$$V{'_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{b} + \frac{{( - 3Q)}}{b}} \right]$
$\therefore $$V{'_{{\rm{sphere }}}} - V{'_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} - \frac{Q}{b}} \right] = V$
${V_{{\rm{inside}}}} = {V_{{\rm{centre }}}} = {V_{{\rm{surface}}}} = \frac{1}{{4\pi {\varepsilon _o}}}.\frac{q}{R}$, ${V_{{\rm{outside}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{r}$
If $a$ and $b$ are the radii of sphere and spherical shell respectively, then potential at their surface will be
${V_{{\rm{sphere }}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{a}$ and ${V_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{b}$
$\therefore $$V = {V_{{\rm{sphere }}}} - {V_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{Q}{a} - \frac{Q}{b}} \right]$
Now when the shell is given charge $(-3Q)$, then the potential will be
$V{'_{{\rm{sphere}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} + \frac{{( - 3Q)}}{b}} \right],$$V{'_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{b} + \frac{{( - 3Q)}}{b}} \right]$
$\therefore $$V{'_{{\rm{sphere }}}} - V{'_{{\rm{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} - \frac{Q}{b}} \right] = V$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A condenser has a capacity $2\,\mu \,F$ and is charged to a voltage of $50\, V$. The energy stored isView Solution
- 2View SolutionA parallel plate capacitor is charged by connecting it to a battery. The battery is disconnected and the plates of the capacitor are pulled apart to make the separation between the plates twice. Again the capacitor is connected to the battery (with same polarity) then
- 3There exists a uniform electric field $E=4 \times 10^5 \,Vm ^{-1}$ directed along negative $x$-axis such that electric potential at origin is zero. Acharge of $-200 \,\mu C$ is placed at origin, and a charge of $+200 \,\mu C$ is placed at $(3 \,m , 0)$. The electrostatic potential energy of the system is ...........$J$View Solution
- 4The total charge on the system of capacitance $C _{1}=1\,\mu F , C _{2}=2\,\mu F , C _{3}=4\,\mu F$ and $C _{4}=3\,\mu F$ connected in parallel is $......\mu C$ (Assume a battery of $20\,V$ is connected to the combination)View Solution
- 5A negatively charged plate has charge density of $2 \times {10^{ - 6}}\,C/{m^2}$. The initial distance of an electron which is moving toward plate, cannot strike the plate, if it is having energy of $200\,eV$View Solution
- 6Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will beView Solution
- 7A particle of charge $Q$ and mass $M$ moves in a circular path of radius $R$ in a uniform magnetic field of magnitude $B$. The same particle now moves with the same speed in a circular path of same radius $R$ in the space between the cylindrical electrodes of the cylindrical capacitor. The radius of the inner electrode is $R/2$ while that of the outer electrode is $ 3R/2.$ Then the potential difference between the capacitor electrodes must beView Solution
- 8A parallel plate capacitor having capacitance $12\, pF$ is charged by a battery to a potential difference of $10\, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$View Solution
- 9A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the closest distances of approach would beView Solution
- 10A series combination of three capacitors of capacities $1\,\mu \,F,\,2\,\mu \,F$ and $8\,\mu \,F$ is connected to a battery of $e.m.f.$ $13\, volt$. The potential difference across the plates of $2\,\mu \,F$ capacitor will be.....$V$View Solution