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Redistribution of Charges Between Concentric Shells

Q: A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are :

For a conducting sphereElectric fidld at centre \$= 0\$Potential at centre \$=\dfrac {KQ}{R}=\dfrac {Q}{4\pi \epsilon_0R}\$

Q: Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force \$F\$, when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

\$\textbf{Step 1: Calculation of initial force}\$ Let \$B\$ and \$C\$ have charge \$Q\$ each initially and are seperated by \$r\$ distance.From Coulomb's law, force between them \$F = \dfrac{K Q^2}{r^2}\$ \$....(1)\$\$\textbf{Step 2: Distribution of charges}\$When an identical uncharged spherical conductor \$A\$ is brought in contact with charged sphere \$B\$.By symmetry, the total charge will be shared equally among them, as both have equal radii.\$\therefore\$ Charge on \$A \$ and \$B\$ \$ = \dfrac{Q}{2}\$ Now, when the conductor \$A\$ is brought in contact with \$C\$. They will also share equal charge among themselves, as both have equal radii.\$\therefore\$ Charge on \$A\$ and \$C\$ \$ = \dfrac{Q + \dfrac{Q}{2}}{2} = \dfrac{3Q}{4}\$\$\textbf{Step 3: Calculation of new force } \$ New coulomb force between \$B\$ and \$C\$:\$F' = \dfrac{K \dfrac{Q}{2} \times \dfrac{3Q}{4}}{r^2} =\dfrac{3}{8} \dfrac{KQ^2}{r^2}\$From equation \$(1)\$, we get: \$F' = \dfrac{3}{8}F\$Hence, Option D is correct

Q: Two concentric spherical conducting shells of radii R and 2R carry charges Q and 2Q respectively, change in electric potential on the outer shell when both are connected by a conducting wire \$\left ( k=\frac{1}{4\pi \epsilon _{0}} \right )\$ is:

The potential due to a charged spherical sphere with radius \$r\$ is given by\$V = \frac{1}{4 \pi \varepsilon_o} \frac {q}{r}= k \frac {q}{r}\$If we two spherical conducting shells of radii \$R\$ and \$2R\$ carry charges \$Q\$ and \$2Q\$, then\$V_1 = k \frac {Q}{2R} \$+ \$k \frac {2 Q}{2R} = k \frac{3Q}{2R}\$Charge remains constant before and after connecting them , so Final Charge \$=3Q\$\$V_2 = k \frac {3Q}{2R} \$Hence \$\Delta V = V_2 - V_1 = 0\$

Q: 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 \$-3Q\$, the new potential difference between the same two surfaces is :

The electric field in between the shell and sphere is \$\displaystyle E.\pi x^2=\frac{Q_{en}}{\epsilon_0}=\frac{Q}{\epsilon_0}\$ (using Gauss's law)\$\displaystyle E=\frac{Q}{4\pi\epsilon_0 x^2}\$The potential difference between the shells is \$\displaystyle dV=V_r-V_R=\int^R_rEdx=\int^R_r\frac{Q}{4\pi\epsilon_0x^2}dx=\frac{Q}{4\pi\epsilon_0}(1/r-1/R)\$Thus, \$V=\frac{Q}{4\pi\epsilon_0}(1/r-1/R)\$As the potential difference between solid sphere and hollow shell depends on the radii of two spheres and charge on the inner sphere, Since the two values have not changed, potential difference does not change. Hence the potential difference remains \$V\$.

Q: Which one of the following graphs shows the variation of electric field strength \$E\$ with distance \$r\$ from the centre of a hollow conducting sphere?

In a uniformly charged conducting sphere, electric field inside \$E=0\$ No charge inclose inside the conductor.Outside \$ E= \dfrac{kq}{r^2}\$Outside it decreases uniformly, so graph(a) is correct.

Q: The electric field at a distance \$\displaystyle \dfrac{3R}{2}\$ from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance \$\displaystyle \dfrac{R}{2}\$ from the centre of the sphere is :

Electric field inside a charged conductor is always zero.

Q: A point charge \$q\$ is placed inside a conducting spherical shell of inner radius \$2R\$ and outer radius \$3R\$ at a distance of \$R\$ from the center of the shell. Find the electric potential at the center of the shell.

Electric potential due to point charge \$=k \dfrac{q}{R}\$ Electric potential due to inner surface of shell \$=-\dfrac{kq}{2R}\$ and electric potential due to outer surface of shell \$=\dfrac{kq}{3R} \$ \$\displaystyle V = k \dfrac{q}{R}-\dfrac{kq}{2R} +\dfrac{kq}{3R} = \dfrac{kq}{R} \left[1-\dfrac{1}{2}+\dfrac{1}{3}\right]\$ \$\displaystyle = \dfrac{k5q}{6R}\$

Q: How many electrons should be removed from a conductor so that it acquires a positive charge of 3.5nC?

Given Charge \$q=3.5\times 10^{-9}\$Let the \$nQ\$ of electron to be removed be \$n\$We know that \$Q=ne\$\$n=\dfrac{Q}{e}\$\$n=\dfrac{3.5\times 10^{-9}}{1.6\times 10^{-19}}\$\$n=2.1875\times 10^{-9+19}\$\$n=2.1875\times 10^{10}\$

Q: A thin metallic spherical shell contains a charge \$Q\$ on its surface. A point charge \$q_{1}\$ is placed at the centre of the shell and another charge \$q_{2}\$ is placed outside the shell. All the three charges are positive. Then, the force on charge \$q_{1}\$ is

\$\textbf{Step 1: Electrostatic Shielding}\$Due to Electrostatic shielding, The metallic spherical shell acts as a shield and do not allow the electric field due to outside charges to penetrate through it.Hence, electric field inside shell due to \$Q\$ and \$q_1\$ is zero.Therefore, no force will act on charge \$q\$.Hence, Correct answer is \$(D)\$\$\textbf{Note : }\$ Electrostatic shielding acts in case of closed metallic surfaces with cavity in which external field does not effect internal contents and internal field does not effect external contents.

Q: Why should electrostatic field be zero inside a conductor?

\$\textbf{Explanation}\$\$\bullet\$The net charge inside a conductor remains zero and the total charge of a conductor resides on its surface as charges want to attain equilibrium so they come on the surface to minimize the repulsion among them.\$\bullet\$As charge inside a conductor is zero so according to gauss law \$\int E.ds=\frac{q}{\epsilon}\$As q=0 \$\implies E=0\$ So the electric field inside the conductor is zero.

Physics

example

Redistribution of charges for concentric spherical shells where one shell is grounded

Two concentric spheres of radius R and 2R have charges Q and 2Q. The potential at a point P situated at a point 3R/2 distance from common centre is V, Now if outer sphere is earthed, the potential at point P is:
Before earthing,the potential at P,

V_{p} = K [\frac{2 Q}{3 R} + \frac{2 Q}{2 R}] = V

so,

\frac{5}{3} K Q = V \Rightarrow K Q = \frac{3 V}{5}

After earthing the outer sphere, potential at P,

V^{'} = K [\frac{2 Q}{3 R} - \frac{Q}{2 R}] = \frac{K Q}{6 R} = \frac{3 V}{5} \times \frac{1}{6 R} = \frac{V}{1 0}

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Important Questions

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are :

Medium

NEET

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Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force $F$ , when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

Hard

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Two concentric spherical conducting shells of radii R and 2R carry charges Q and 2Q respectively, change in electric potential on the outer shell when both are connected by a conducting wire $(k = \frac{1}{4 π ϵ _{0}})$ is:

Medium

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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 $- 3 Q$ , the new potential difference between the same two surfaces is :

Hard

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Which one of the following graphs shows the variation of electric field strength $E$ with distance $r$ from the centre of a hollow conducting sphere?

Easy

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The electric field at a distance $\frac{3 R}{2}$ from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance $\frac{R}{2}$ from the centre of the sphere is :

Medium

NEET

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A point charge $q$ is placed inside a conducting spherical shell of inner radius $2 R$ and outer radius $3 R$ at a distance of $R$ from the center of the shell. Find the electric potential at the center of the shell.

Medium

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How many electrons should be removed from a conductor so that it acquires a positive charge of 3.5nC?

Easy

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A thin metallic spherical shell contains a charge $Q$ on its surface. A point charge $q_{1}$ is placed at the centre of the shell and another charge $q_{2}$ is placed outside the shell. All the three charges are positive. Then, the force on charge $q_{1}$ is

Medium

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Why should electrostatic field be zero inside a conductor?

Medium

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example

Redistribution of charges for concentric spherical shells where one shell is grounded

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Redistribution of the Charges in Three Concentric Spherical Shells

A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the centre of the sphere respectively are :

Two concentric spherical conducting shells of radii R and 2R carry charges Q and 2Q respectively, change in electric potential on the outer shell when both are connected by a conducting wire (k=4πϵ0​1​) is:

Which one of the following graphs shows the variation of electric field strength E with distance r from the centre of a hollow conducting sphere?

The electric field at a distance 23R​ from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance 2R​ from the centre of the sphere is :

A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the center of the shell. Find the electric potential at the center of the shell.

How many electrons should be removed from a conductor so that it acquires a positive charge of 3.5nC?

A thin metallic spherical shell contains a charge Q on its surface. A point charge q1​ is placed at the centre of the shell and another charge q2​ is placed outside the shell. All the three charges are positive. Then, the force on charge q1​ is

Why should electrostatic field be zero inside a conductor?

Two concentric spherical conducting shells of radii R and 2R carry charges Q and 2Q respectively, change in electric potential on the outer shell when both are connected by a conducting wire $(k = \frac{1}{4 π ϵ _{0}})$ is:

Which one of the following graphs shows the variation of electric field strength $E$ with distance $r$ from the centre of a hollow conducting sphere?

The electric field at a distance $\frac{3 R}{2}$ from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance $\frac{R}{2}$ from the centre of the sphere is :

A point charge $q$ is placed inside a conducting spherical shell of inner radius $2 R$ and outer radius $3 R$ at a distance of $R$ from the center of the shell. Find the electric potential at the center of the shell.

A thin metallic spherical shell contains a charge $Q$ on its surface. A point charge $q_{1}$ is placed at the centre of the shell and another charge $q_{2}$ is placed outside the shell. All the three charges are positive. Then, the force on charge $q_{1}$ is