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: 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: 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.

Question 1

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 :

Accepted Answer

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

Question 2

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:

Accepted Answer

$\textbf{Step 1: Calculation of initial force}$ 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

Question 3

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:

Accepted Answer

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 $\Delta V = V_2 - V_1 = 0$

Question 4

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 :

Accepted Answer

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 $V$.

Question 5

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 :

Accepted Answer

Electric field inside a charged conductor is always zero.

Question 6

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?

Accepted Answer

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.

Question 7

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.

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Electric potential due&#10;to point charge $=k \dfrac{q}{R}$&#10;&#10;Electric potential due&#10;to inner surface of shell $=-\dfrac{kq}{2R}$&#10;&#10;and electric potential&#10;due to outer surface of shell $=\dfrac{kq}{3R} $&#10;&#10;$\displaystyle V = k&#10;\dfrac{q}{R}-\dfrac{kq}{2R} +\dfrac{kq}{3R} = \dfrac{kq}{R} \left[1-\dfrac{1}{2}+\dfrac{1}{3}\right]$&#10;&#10;$\displaystyle = \dfrac{k5q}{6R}$&#10;&#10;

Question 8

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

Accepted Answer

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}$

Question 9

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

Accepted Answer

$\textbf{Step 1: Electrostatic Shielding}$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.

Question 10

Why should electrostatic field be zero inside a conductor?

Accepted Answer

$\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.

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Electric Field and Potential Due to Conductor

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