Coulomb’s Law of Electrostatics Notes

Electrostatics Synopsis

  • An electric charge (q or Q) is always associated with electrons and protons which are constituents of atoms.
  • The SI unit of charge is the coulomb (symbol: C). The elementary charge is e = 1.6 x 10-19 C.
  • 1 C of charge is contained in 6.25 x 1018 electrons (or protons).
  • A charge is quantized as Q = ne, where n = 1, 2, 3,…
  • A charging process involves a gain or loss of electrons. An object becomes positively charged due to the loss of electrons and negatively charged due to the gain of electrons.
  • Coulomb’s law: \(F=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{r^2}\right)=K\left(\frac{q_1 q_2}{r^2}\right)\)
    where ε0 = permittivity of free space
    = 8.85 x 10-12 F m-1 (or C2 N-1 m-2)
    and \(K=\frac{1}{4 \pi \varepsilon_0}\)
  • = \(9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\)
  • Principle of superposition: The net force on any charge q due to other charges in space equals the vector sum of all the forces on it on account of the constituent charges in space. Thus,
    ⇒ \(\vec{F}=\vec{F}_1+\vec{F}_2+\ldots=\frac{1}{4 \pi \varepsilon_0}\left[q\left(\frac{q_1}{r_1^2} \cdot \hat{r}_1+\frac{q_2}{r_2{ }^2} \cdot \hat{r}_2+\ldots\right)\right]\)
  • The electric field (or field strength or field intensity) \(\vec{E}\) is defined as the force per unit test charge \(\left(\vec{E}=\frac{\vec{E}}{q_0}\right)\). Its SI unit is NC-1(≡V m-1).
    The electric field due to a single charge Q (monopole) is
    ⇒ \(\vec{E}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{r^2} \cdot \hat{r}\right)\)
  • Electric dipole: Two equal and opposite point charges separated at a short distance constitute an electric dipole.
  • The electric dipole moment \((\vec{p})\) is the product of one charge |± q| and the separation (δl) between the charges. Thus,
    ⇒ \(|\vec{p}|=|q| \delta l \Rightarrow \vec{p}=|q| \vec{\delta} l\)
    Here \(\vec{p}\) is a vector directed from the negative charge to the positive charge, and its SI unit is the coulomb meter (C m).
  • Dipole field:
    • At a point on the axis,
      ⇒ \(\vec{E}_{\text {axial }}=\frac{1}{4 \pi \varepsilon_0}\left[\frac{2 \vec{p} r}{\left(r^2-l^2\right)^2}\right] \approx \frac{1}{4 \pi \varepsilon_0}\left(\frac{2 \vec{p}}{r^3}\right)\)
    • At an equatorial point,
      ⇒ \(\vec{E}_{\text {equatorial }}=\frac{1}{4 \pi \varepsilon_0}\left[\frac{-\vec{p}}{\left(r^2+l^2\right)^{3 / 2}}\right] \approx \frac{1}{4 \pi \varepsilon_0}\left(\frac{-\vec{p}}{r^3}\right)\)
    • At any point P(r, θ),
      ⇒ \(E=\frac{1}{4 \pi \varepsilon_0}\left(\frac{p}{r^3} \sqrt{1+3 \cos ^2 \theta}\right)\)
  • Torque on a dipole in a uniform field \(\vec{E}\):
    ⇒ \(\vec{\tau}=\vec{p} \times \vec{E} \Rightarrow \tau=p E \sin \theta\)
  • Flux of an electric field:
    ⇒ \(\phi=\int \vec{E} \cdot \overrightarrow{d A}=\int E d A \cos \theta\)
    Φ is a scalar quantity, and its SI unit is V m (≡ N m2 C-1).
  • Gauss’s law: \(\Phi=\frac{1}{\varepsilon_0}\) (net charge enclosed).
  • Expressions for an electric field:
    • For a line charge, \(\vec{E}=\frac{\lambda}{2 \pi \varepsilon_0 r} \cdot \hat{r}\)
    • For a plane charge sheet, \(\vec{E}=\frac{\sigma}{2 \varepsilon_0} \cdot \hat{n}\)
    • For a uniformly charged spherical shell of radius R,
      ⇒ \(\vec{E}=\left\{\begin{array}{cc}
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{r^2} \cdot \hat{r}\right) & \text { for } r>R \\
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R^2} \cdot \hat{R}\right) & \text { for } r=R \\
      0 & \text { for } r<R
      \end{array}\right.\)
    • For a solid sphere of radius R having a total charge of Q distributed uniformly,
      ⇒ \(\vec{E}=\left\{\begin{array}{l}
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{r^2} \cdot \hat{r}\right) \text { for } r>R \\
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R^2} \cdot \hat{R}\right) \text { for } r=R \\
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q r}{R^3} \cdot \hat{r}\right) \text { for } r<R
      \end{array}\right.\)
  • Potential energy between two charges:
    ⇒ \(U=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{r}\right)\)
  • The potential energy of a system of three charges:
    ⇒ \(U_{\text {sys }}=U_{12}+U_{13}+U_{23}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{r_{12}}+\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right)\)
  • Potential at a point in an electric field:
    V = \(\frac{U}{q}\) = potential energy per unit test charge.

    • The potential due to a monopole (single charge) Q is
      ⇒ \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{r}\right)\)
    • The potential due to a system of charges is
      ⇒ \(V=V_1+V_2+\ldots=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q_1}{r_1}+\frac{Q_2}{r_2}+\ldots\right)\)
    • The potential due to a uniformly charged ring of radius R is
      ⇒ \(V=\left\{\begin{array}{l}
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{\sqrt{R^2+r^2}}\right) \text { atP } \\
      \frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}\right) \text { atO }
      \end{array}\right.\)

Electrostatics Synopsis The potential due to a uniformly changed ring of radius R

  • Potential due to an electric dipole:
    • At an axial point, V = \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{p}{r^2}\right)\)
    • At an equatorial point, V = 0.
    • At any point \(\mathrm{P}(r, \theta), V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{p \cos \theta}{r^2}\right)\)
  • Potential due to a uniformly charged spherical shell of radius R:
    • At an external point, \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{r}\right)\)
    • On the surface of the shell, \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}\right)\)
    • Inside theshell, \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}\right)\) (the same as thaton thesurface).
  • Potential due to a spherical charge distribution:
    • At an external point, \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{r}\right) \text { for } r>R\)
    • On the surface of the sphere, \(V=\frac{1}{4 \pi \varepsilon_0}\left(\frac{Q}{R}\right)\)
    • Inside the sphere at a distance r (where r < R) from the center,
      ⇒ \(V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q\left(3 R^2-r^2\right)}{2 R^3}\right]\)
  • The potential energy of an electric dipole in a uniform electric field \(\vec{E}\):
    ⇒ \(U=-\vec{p} \cdot \vec{E}=-p E \cos \theta\)
  • The potential difference in an electric field:
    ⇒ \(V_2-V_1=-\int_1^2 \vec{E} \cdot \vec{d} r\)
  • Relation between the field \(\vec{E}\) and the potential (V):
    ⇒ \(|\vec{E}|=-\frac{d V}{d r}, \text { where } E_x=-\frac{\partial V}{\partial x}, E_y=-\frac{\partial V}{\partial y} \text { and } E_z=-\frac{\partial V}{\partial z}\)
  • The electric field in the material of a conductor is \(\vec{E}\) = 0, and that in a dielectric medium of dielectric constant K is \(\vec{E}=\frac{\vec{E}_0}{K} \text {, where } \vec{E}_0\) is the external field strength.
  • Dielectric strength: A sufficient increase in the external electric field reduces the ability of atoms and molecules to hold the outer electrons, which get detached to cause a dielectric breakdown. The maximum electric field that- the dielectric can sustain without breaking down is called its dielectric strength. For dry air, the dielectric strength is around 3 x 106 V m-1.
  • Electrical capacitance, C = \(\frac{Q}{V}\). Its SI unit is the farad (symbol F).
    Thus,
    1F = \(\frac{1C}{1V}\)
  • The capacitance of a sphere of radius R is C = 4nε0R
  • Electrostatic energy, \(U=\frac{1}{2} C V^2=\frac{Q^2}{2 C}\)
  • The capacitance of a parallel-plate capacitor:
    • With air, \(C_0=\frac{\varepsilon_0 A}{d}\)
    • With a dielectric, \(C=K C_0=\frac{K \varepsilon_0 A}{d}\)
    • With a thin dielectric plate, \(C=\frac{\varepsilon_0 A}{d-t+\frac{t}{K}} \text { for } t<d\)
    • With a thin metal plate, \(C=\frac{\varepsilon_0 A}{d-t} \text { for } t<\underline{d} \text { and } K_{\text {metal }}=\infty\)
  • Equivalent capacitance:
    • In series, \(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\ldots\)
    • In parallel, Cp = C1 + C2 + ….
      Cs is less than the least in the combination, Cp is greater than the greatest in the combination.

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