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.

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

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}\\)

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\\).

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\\)

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.

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}\\)

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.

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}\\)

A hollow insulated conducting sphere is given a positive charge of \\(10\mu C\\). What will be the electric field at the centre of the sphere if its radius is \\(2\\) metres?

\\(\textbf{Step 1: Uniform Charge distribution on outer surface [Refer Figure]}\\) As a property of conductor, Charge will resides on the outer surface of the hollow sphere.Since the shape of sphere is symmetric, hence charge distribution will be uniform, As shown in the figure.\\(\textbf{Step 2: Finding Electric field inside}\\)The given situation is now a uniformly charged hollow conducting shell.Inside which the Electric field is zero at all points.Hence correct option is \\(A\\).\\(\large \textbf{Alternate Solution using Gauss Law:}\\)We can find the electric field inside using Gauss Law as follows:Consider a gaussian spherical surface of radius \\(r<R\\)Charge inside gaussian surface \\(q_{in}=0\\)From gauss theorem:\\(\oint _{ }^{ }{ \vec { E } .\vec {ds} } \\) \\(=\cfrac { q_{in} }{ { \varepsilon }_{ 0 } } \\) At all points of gaussian surface, \\(\vec E\\) is constant radially outwards due to symmetry and \\(\vec {ds}\\) is perpendicular to the gaussian surface, hence both are parallel and angle between them is zero.\\(\therefore\ \ \ \vec E \oint _{ }^{ }{ \vec {ds}\ cos0^{o}} \\) \\(= \vec E \oint \vec {ds}\\) \\(=\cfrac { q_{in} }{ { \varepsilon }_{ 0 } } \\) Full area of gaussian surface is \\(4 \pi r^2\\)\\(\Rightarrow\ \ \ \ E\times 4\pi { r }^{ 2 }=\cfrac { 1 }{ { \varepsilon }_{ 0 } } \times 0\quad \\)\\(\Rightarrow \quad E=0\\)Hence, option A is correct.

Electric Field and Potential Due to Conductor | Definition, Examples, Diagrams

definition

Electrostatic shielding

Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence: the field inside the cavity is always zero. This is known as electrostatic shielding.

definition

Electrostatic field is zero inside a conductor

Consider a conductor, neutral or charged. There may also be an external electrostatic field.
In the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. This fact can be taken as the defining property of a conductor.
A conductor has free electrons. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.

definition

At the surface of a charged conductor, electrostatic field must be normal to the surface at every point

If E were not normal to the surface, it would have some non-zero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore,E should have no tangential component. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface.)

definition

The interior of a conductor can have no excess charge in the static situation

A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. When the conductor is charged,the excess charge can reside only on the surface in the static situation.This follows from the Gauss's law. Consider any arbitrary volume element v inside a conductor. On the closed surface S bounding the volume element v, electrostatic field is zero. Thus the total electric flux through S is zero. Hence, by Gauss's law, there is no net charge enclosed by S. But the surface S can be made as small as you like, i.e., the volume v can be made vanishingly small. This means there is no net charge at any point inside the conductor, and any excess charge must reside at the surface.

definition

Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface

Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor. Hence, the result. If the conductor is charged,electric field normal to the surface exists; this means potential will be different for the surface and a point just outside the surface.In a system of conductors of arbitrary size, shape and charge configuration, each conductor is characterised by a constant value of potential, but this constant may differ from one conductor to the other.

definition

Electric field at the surface of a charged conductor

E = \frac{σ}{ϵ _{0}} \overset{n}{^}

where

σ

is the surface charge density and

\overset{n}{^}

is a unit vector normal to the surface in the outward direction.
To derive the result, choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface. The pillbox is partly inside and partly outside the surface of the conductor. It has a small area of cross section

δ S

and negligible height.Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E. Thus,the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box. This equals

\pm E δ S

(positive for

σ > 0

,negative for

σ < 0

), since over the small area

δ S

E

may be considered constant and

E

and

δ S

are parallel or antiparallel. The charge enclosed by the pill box is

σ δ S

.By Gauss's law

E δ S = \frac{∣ σ ∣ δ S}{ϵ _{0}}

E = \frac{∣ σ ∣}{ϵ _{0}}

definition

Describe the uses of Van de Graaff Generator

Small Van de Graaff machines are produced for entertainment, and in physics education to teach electrostatics; larger ones are displayed in science museums.

Electric Field and Potential Due to Conductor

definition

Electrostatic shielding

definition

Electrostatic field is zero inside a conductor

definition

At the surface of a charged conductor, electrostatic field must be normal to the surface at every point

definition

The interior of a conductor can have no excess charge in the static situation

definition

Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface

definition

Electric field at the surface of a charged conductor

definition

Describe the uses of Van de Graaff Generator

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