We will discuss the essential points to be remembered in electrostatics under electric field and potential. Gauss’s law as well as field and potential due to charge configurations other than those involving point charges are meant for AP Physics C only.

**(1)**

**: The electrostatic force between two point charges**

*Coulomb’s Law**q*

_{1}and

*q*

_{2 }is proportional to the product

*q*

_{1}

*q*

_{2}and inversely proportional to the square of the distance

*r*

_{ }between them.

This is expressed mathematically as,

*F*=

*k*(

*q*

_{1}

*q*

_{2})

_{ }/

*r*

^{2}

where the constant

*k**=*1/4πε_{0}which is very nearly equal to 9×10^{9}Nm^{2}C^{–2}. The constant ε_{0}is the permittivity of free space. The equation for electrostatic force when the charges are in free space is therefore written as*F*= (1/4πε

_{0})

*(*

*q*

_{1}

*q*

_{2})

_{ }/

*r*

^{2}

[If the charges are placed in a medium of relative permittivity ε

_{r}, the above equation for electrostatic force will become*F*= (1/4πε_{0}ε_{r})*(**q*_{1}*q*_{2})_{ }/*r*^{2}].
In vector form, the equation for electrostatic force when the charges are in free space is

**F**

_{21}=

*k*(

*q*

_{1}

*q*

_{2})

_{ }

**ř**

**/**

_{21}

*r*

_{21}

^{2}

^{}
In this form

**F**_{21}_{ }is the force (vector) on*q*_{2}due to*q*_{1},**ř**_{21}**is a unit vector in the direction from**_{ }*q*_{1}to*q*_{2}_{ }and*r**is the distance between*_{21}*q*_{1}and*q*_{2}.**(2)**

**The electric field**

**E**

**at any point is the force on a**

*small positive*test charge

*q*placed at the point divided by the magnitude of the test charge:

**E**=

**F**/

*q*

**(3)**

**Electric field**

**(**

*magnitude*

*E*

**)**

**due to a**

*point charge*

*Q***at distance**

*r*is given by

*E*= (1/4πε

_{0})

*Q/*

*r*

^{2}. It is radially outwards from

*q*, if

*q*is positive and radially inwards if

*q*is negative.

**(4)**

**Electric field due to a**

**n electric dipole**

**at a point on its axis**distant

*r*from the centre of the dipole s given by

**E**

*=*(1/4πε

_{0}) 2

**p**

*r*/(

*r*

^{2 }–

*a*

^{2})

^{2}where

**p**is the dipole moment vector of magnitude 2

*qa*directed from –

*q*to +

*q*.

[You should remember that 2

*a*is the separation between the charges –*q*and +*q*constituting the dipole].
The

*magnitude*of the electric field due to an electric dipole at a point on its axis distant*r*from the centre of the dipole s given by*E*

*=*(1/4πε

_{0}) 2

*p*

*r*/(

*r*

^{2 }–

*a*

^{2})

^{2}where

*p =*2

*qa*.

At a distance

*r**large*compared to the dipole length 2*a*, the field is*E*

*=*(1/4πε

_{0}) 2

*p*/

*r*

^{3}

**(5**

**)**

**Electric field due to a**

**n electric dipole**

**at a point**

**in its equatorial plane**(i.e., the plane perpendicular to its axis and passing through its centre) distant

*r*from the centre of the dipole s given by

**E**

*=*(1/4πε

_{0}) (–

**p**

**)**/(

*r*

^{2 }+

*a*

^{2})

^{3/2}

The field in the equatorial plane is antiparallel to the dipole moment vector as indicated by –

**p****in the above expression.**
The

*magnitude*of the electric field due to an electric dipole at a point in its equatorial plane distant*r*from the centre of the dipole is given by*E*

*=*(1/4πε

_{0})

*p*/

*r*

^{3}

[Note that the expression for the electric field due to an electric dipole is similar to the expression for the magnetic field due to a magnetic dipole, which is obtained by replacing (1/4πε

_{0}) with (μ_{0}/4π) and the electric dipole moment**p**with the magnetic dipole moment**m**].**(6)**Torque (

**τ**) on an electric dipole of moment

**p**placed in a uniform electric field

**E**is given by

**τ**

**=**

**p**

**×E**

Therefore,

*τ***=***pE*sin*θ*where*θ*is the angle between**p**and**E**
[Remember that the dipole moment vector has magnitude 2

*qa*and its direction is from –*q*to +*q*].**(7)**

**Gauss’s Law (Gauss Theorem):**The flux (

*Ф*) of electric field through any closed surface

*S*is 1/ε

_{0}times the total charge enclosed by the surface

*S*.

*Ф*= ∫E.dS = Q

*/*ε

_{0}where the integration is over the closed surface

*S*which encloses a total charge Q.

**(8)**

**Electric fields due to some symmetric charge configurations**:

**(**

**i**

**)**

**Electric field**

**(**

*magnitude*

*E*

**)**

**due to a**

**carrying charge**

*spherical conductor*

**at a point**

*Q**outside*the sphere at distance

*r*from the centre of the sphere is given by

*E*= (1/4πε

_{0})

*Q/*

*r*

^{2}.

The field on the surface of the spherical conductor is

*E*= (1/4πε

_{0})

*Q/R*

^{2}where

*R*is the radius of the sphere.

As far as points

*outside**the sphere and on the sphere*are concerned, the electric field due to a charged spherical*conductor*has values as though the entire charge is concentrated at the centre.**(ii)**

**Electric field**

**at any point**

*inside a conductor***carrying charge**

*Q*

**is**

*zero***since the charge resides only on the surface of the conductor.**

**(**

**iii**

**)**

**Electric field**

**(**

*magnitude*

*E*

**)**

**due to a**

**uniformly charged**

**thin**

**carrying charge**

*spherical shell*

*Q*

**is given by**

*E*= (1/4πε

_{0})

*Q/*

*r*

^{2}at a point distant

*r*from the centre,

*outside the shell.*

On the surface of the shell,

*E*= (1/4πε_{0})*Q/R*^{2}where*R*is the radius of the shell.
Obviously, at any point

*inside a**uniformly charged thin spherical**shell*carrying charge*Q**,***the field****is****.***zero***(iv)**

**Electric field due to a**

*spherical distribution of charges with a uniform volume charge density***:**

*ρ*
At a point

*outside*the sphere at distance*r*from the centre of the sphere the field is given by*E*=

*ρR*

^{3}

*/*3ε

_{0}r

^{2}), which follows by substituting

*Q*= (4/3) π

*R*

^{3}

*ρ*in the expression

*E*= (1/4πε

_{0})

*Q/*

*r*

^{2}.

At a point

*o**n*t*he**surface of the**sphere*the field is given by*E*=

*ρR/*3ε

_{0}, which follows by substituting

*r*=

*R*in the above expression.

At a point

*in**side*the sphere at distance*r*from the centre of the sphere the field is given by*E*=

*ρr/*3ε

_{0}), which follows by substituting

*Q*= (4/3) π

*r*

^{3}

*ρ*in the expression

*E*= (1/4πε

_{0})

*Q/*

*r*

^{2}.

**(v)**

**Electric**

**field**at distance

*r*from an infinitely long

**with linear charge density λ is**

*straight uniformly charged wire**E*= λ/2πε

_{0}r

The field is perpendicular to the wire.

**(v**

**i**

**)**

**Electric**

**field**due to a uniformly charged

**with uniform surface charge density σ:**

*infinite plane sheet**E*=

*σ*/2ε

_{0}, which is independent of the distance of the point from the surface.

The field is directed normal to the surface.

**(9) Electrostatic p**

**otential**(

*V*) at a point is the work done (by an external agency) in bringing unit positive charge from infinity to that point.

By convention, the potential at infinity is taken as zero.

**(i)**

**Electr**

**ostat**

**ic**

**potential**

**due to a**

*point charge*

*Q***at distance**

*r*is given by

*V =*(1/4πε

_{0})

*Q/*

*r*

**(ii**

**)**

**Potential at a point P due to a system of charges**

*q*

_{1},

*q*

_{2},

*q*

_{3},….etc. at distances

*r*

_{1},

*r*

_{2},

*r*

_{3},….etc. is given by

*V =*(1/4πε

_{0})[(

*q*

_{1}

*/*

*r*

_{1}) + (

*q*

_{2}

*/*

*r*

_{2}) + (

*q*

_{3}

*/*

*r*

_{3}) +…etc.]

**(iii)**

**Electr**

**ostat**

**ic**

**potential**

**due to**

**an**

*electric dipole***at distance**

*r*(which is very large compared to the dipole length 2a) from the centre of the dipole is given by

*V =*

*p*

*cos*

*θ*/4πε

_{0}

*r*

^{2}

^{}
where

*p*is the magnitude of the dipole moment and*θ*is the angle between the dipole moment vector and the line joining the dipole to the point P where the potential is measured.
[Note that potential is a scalar quantity. In terms of the dipole moment vector

**p**and the position vector**r**of the point P, taking the centre of the dipole as the origin, the potential of the dipole can be written as*V*=*=***p****.r**/4πε_{0}*r*^{3}. Or,*V*=*=***p****.****ř**/4πε_{0}*r*^{2}where**ř**is a unit vector along the position vector**r**].**(iv)**

**Electric**

**potential**

**due to a**

*spherical conductor***of radius**

*R*

**carrying charge**

*Q*

**:**

At a point

*outside*the sphere at distance*r*from the centre of the sphere, the potential is given by

*V =*(1/4πε

_{0})

*Q/*

*r*

At a point

*o**n*t*he**surface of the**sphere*the potential is given by*V =*(1/4πε

_{0})

*Q/*

*R*

At

*all points**in**side*the sphere, the potential is the same as the value at the surface given by*V =*(1/4πε

_{0})

*Q/*

*R*

**(**

**v**

**)**

**Electric**

**potential at any point**

**due to a**

**uniformly charged**

**thin**

**carrying charge**

*spherical shell*

*Q*

**is the same as for a charged spherical conductor (given above).**

**(10) Electrostatic potential energy**of a charge

*q*at a point where the electric potential is

*V*is

*Vq*.

**(11) Electrostatic potential energy (**

*U*) of a system of two point charges**is given by**

*q*and_{1}*q*_{2}separated by a distance*r**U =*(1/4πε

_{0})

*q*

_{1}

*q*

_{2}

*/*

*r*

Note that

*U*is the work done (by external agency) in reducing the separation between the charges from infinity to*r*. Its value will be positive or negative, depending on whether the product*q*_{1}*q*_{2}is positive or negative.
If there are more than two charges, there will be as many terms in the expression for electrostatic potential energy as there are independent pairs of charges. Thus, if there are three point charges

*q*_{1},_{ }*q*_{2}and*q*_{3 }there will be three terms:*U =*(1/4πε

_{0})[(

*q*

_{1}

*q*

_{2}

*/r*

_{12}) + (

*q*

_{1}

*q*

_{3}

*/r*

_{13}) + (

*q*

_{2}

*q*

_{3}

*/r*

_{23})] where

*r*

_{12}is the distance between charges

*q*

_{1 }and

*q*

_{2},

*r*

_{13}is the distance between charges

*q*

_{1 }and

*q*

_{3}and

*r*

_{23}is the distance between charges

*q*

_{2 }and

*q*

_{23}.

If there are four point charges, there will be six terms in

*U*.
We will discuss questions in this section in the next post.

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