Electrostatics Concept Page - 14

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

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.

Definition

Electric field at the surface of a charged conductor

E=σϵ0n^$E={\displaystyle \frac{\sigma }{{ϵ}_0}}\hat{n}$
where σ$\sigma$  is the surface charge density and n^$\hat{n}$ 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 δS$\delta S$ 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 ±EδS$\pm E\delta S$  (positive for σ>0$\sigma >0$,negative for σ<0$\sigma <0$), since over the small area δS$\delta S$, E$E$ may be considered constant and E$E$ and δS$\delta S$ are parallel or antiparallel. The charge enclosed by the pill box is σδS$\sigma \delta S$.By Gauss's law
EδS=|σ|δSϵ0$E\delta S={\displaystyle \frac{|\sigma |\delta S}{{ϵ}_0}}$
E=|σ|ϵ0$E={\displaystyle \frac{|\sigma |}{{ϵ}_0}}$

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

Redistribution of charges for arrangement of parallel plates

Total charge on A is Q1$Q_1$ and on B is Q2$Q_2$ which remains constant by law of conservation of charge.
From gauss's law, total charge on the inner regions is zero. Hence, equal and opposite charges are induced on the inner surface. 
Net electric field at P is zero.
Q1−q2Aεo−q2Aεo+q2Aεo−Q2+q2Aεo${\displaystyle \frac{Q_1-q}{2A{\epsilon }_o}}-{\displaystyle \frac{q}{2A{\epsilon }_o}}+{\displaystyle \frac{q}{2A{\epsilon }_o}}-{\displaystyle \frac{Q_2+q}{2A{\epsilon }_o}}$
q=Q1−Q22$q={\displaystyle \frac{Q_1-Q_2}{2}}$

Definition

Redistribution of charges for arrangement of parallel plates where one plate is grounded

Total charge on A is Q1$Q_1$ and on B is 0 as it is grounded which remains constant by law of conservation of charge.
From gauss's law, total charge on the inner regions is zero. Hence, equal and opposite charges are induced on the inner surface. 
Net electric field at P is zero.
Q1−q2Aεo−q2Aεo+q2Aεo−q2Aεo${\displaystyle \frac{Q_1-q}{2A{\epsilon }_o}}-{\displaystyle \frac{q}{2A{\epsilon }_o}}+{\displaystyle \frac{q}{2A{\epsilon }_o}}-{\displaystyle \frac{q}{2A{\epsilon }_o}}$
q=Q12$q={\displaystyle \frac{Q_1}{2}}$

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, Vp=K[2Q3R+2Q2R]=V$V_p=K[{\displaystyle \frac{2Q}{3R}}+{\displaystyle \frac{2Q}{2R}}]=V$
so, 53KQ=V⇒KQ=3V5${\displaystyle \frac{5}{3}}KQ=V\Rightarrow KQ={\displaystyle \frac{3V}{5}}$
After earthing the outer sphere, potential at P, V′=K[2Q3R−Q2R]=KQ6R=3V5×16R=V10$V^{\prime }=K[{\displaystyle \frac{2Q}{3R}}-{\displaystyle \frac{Q}{2R}}]={\displaystyle \frac{KQ}{6R}}={\displaystyle \frac{3V}{5}}\times {\displaystyle \frac{1}{6R}}={\displaystyle \frac{V}{10}}$

Example

Redistribution of charges for arrangement of spherical conductors

Example:
Two concentric thin conducting spherical shells having radius R$R$ and 2R$2R$ are as shown in figure. A charge +Q$+Q$ is given to shell A and −4Q$-4Q$ is given to shell B. Now shell A and B are connected by a thin conducting wire. Find the distribution of charges after making the connection.
Solution:After connecting both the spherical shells by a conducting wire, they will have the same potential. Let Q1$Q_1$ and Q2$Q_2$ be the charges on the inner sphere and the outer sphere respectively after they are connected.

Potential of the spherical shell having radius R$R$: VR=Q1R+Q22R$V_R={\displaystyle \frac{Q_1}{R}}+{\displaystyle \frac{Q_2}{2R}}$
Potential of the spherical shell having radius 2R$2R$: V2R=Q12R+Q22R$V_{2R}={\displaystyle \frac{Q_1}{2R}}+{\displaystyle \frac{Q_2}{2R}}$
Since VR=V2R$V_R=V_{2R}$
Q1R+Q22R=Q12R+Q22R${\displaystyle \frac{Q_1}{R}}+{\displaystyle \frac{Q_2}{2R}}={\displaystyle \frac{Q_1}{2R}}+{\displaystyle \frac{Q_2}{2R}}$
⇒Q1=0$\Rightarrow Q_1=0$
But Q1+Q2=Q+(−4Q)=−3Q$Q_1+Q_2=Q+(-4Q)=-3Q$
⇒0+Q2=−3Q$\Rightarrow 0+Q_2=-3Q$
⇒Q2=−3Q

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