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) Coulomb’s Law: The electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r between them.
This is expressed mathematically as,
F = k (q1q2) /r2
where the constant k = 1/4πε0 which is very nearly equal to 9×109 Nm2C–2. The constant ε0 is the permittivity of free space. The equation for electrostatic force when the charges are in free space is therefore written as
F = (1/4πε0) (q1q2) /r2
[If the charges are placed in a medium of relative permittivity εr, the above equation for electrostatic force will become F = (1/4πε0εr) (q1q2) /r2].
In vector form, the equation for electrostatic force when the charges are in free space is
F21 = k (q1q2) ř21/ r212
In this form F21 is the force (vector) on q2 due to q1, ř21 is a unit vector in the direction from q1 to q2 and r21 is the distance between q1 and q2.
(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 = (1/4πε0)Q/r2. It is radially outwards from q, if q is positive and radially inwards if q is negative.
(4) 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 = (1/4πε0) 2pr/(r2 – a2)2 where p is the dipole moment vector of magnitude 2qa directed from –q to +q.
[You should remember that 2a 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 = (1/4πε0) 2pr/(r2 – a2)2 where p = 2qa.
At a distance r large compared to the dipole length 2a, the field is
E = (1/4πε0) 2p/r3
(5) Electric field due to an 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 = (1/4πε0) (–p)/(r2 +a2)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 = (1/4πε0) p/r3
[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 (1/4πε0) with (μ0/4π) 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 sinθ where θ is the angle between p and E
[Remember that the dipole moment vector has magnitude 2qa 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/ε0 times the total charge enclosed by the surface S.
Ф = ∫E.dS = Q/ε0 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 spherical conductor carrying charge Q at a point outside the sphere at distance r from the centre of the sphere is given by
E = (1/4πε0)Q/r 2.
The field on the surface of the spherical conductor is
E = (1/4πε0)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 spherical shell carrying charge Q is given by
E = (1/4πε0)Q/r 2 at a point distant r from the centre, outside the shell.
On the surface of the shell, E = (1/4πε0)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 = ρR3/3ε0r2), which follows by substituting Q = (4/3) πR3ρ in the expression E = (1/4πε0)Q/r 2.
At a point on the surface of the sphere the field is given by
E = ρR/3ε0, which follows by substituting r = R in the above expression.
At a point inside the sphere at distance r from the centre of the sphere the field is given by
E = ρr/3ε0), which follows by substituting Q = (4/3) πr3ρ in the expression E = (1/4πε0)Q/r 2.
(v) Electric field at distance r from an infinitely long straight uniformly charged wire with linear charge density λ is
E = λ/2πε0r
The field is perpendicular to the wire.
(vi) Electric field due to a uniformly charged infinite plane sheet with uniform surface charge density σ:
E = σ/2ε0, which is independent of the distance of the point from the surface.
The field is directed normal to the surface.
(9) Electrostatic potential (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) Electrostatic potential due to a point charge Q at distance r is given by
V = (1/4πε0)Q/r
(ii) Potential at a point P due to a system of charges q1, q2, q3,….etc. at distances r1, r2, r3,….etc. is given by
V = (1/4πε0)[(q1/r1) + (q2/r2) + (q3/r3) +…etc.]
(iii) Electrostatic 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 cosθ /4πε0r2
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/4πε0r3. Or, V = = p.ř/4πε0r2 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 = (1/4πε0)Q/r
At a point on the surface of the sphere the potential is given by
V = (1/4πε0)Q/R
At all points inside the sphere, the potential is the same as the value at the surface given by
V = (1/4πε0)Q/R
(v) Electric potential at any point due to a uniformly charged thin spherical shell carrying charge 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 q1 and q2 separated by a distance r is given by
U = (1/4πε0)q1q2 /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 q1q2 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 q1, q2 and q3 there will be three terms:
U = (1/4πε0)[(q1q2 /r12) + (q1q3 /r13) + (q2q3 /r23)] where r12 is the distance between charges q1 and q2, r13 is the distance between charges q1 and q3 and r23 is the distance between charges q2 and q23.
If there are four point charges, there will be six terms in U.
We will discuss questions in this section in the next post.
Meanwhile, find some useful multiple choice questions from electrostatics here as well as here
e, the flux through one pair of opposite faces of the cube is Q/3ε0. 
