Question

(a) Two large conducting spheres carrying charges ${Q_1}$ ​ and ${Q_2}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $\dfrac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$, where $r$ is the distance between their centres?

(b) If Coulomb's law involved $\dfrac{1}{{{r^3}}}$ dependence (instead of $\dfrac{1}{{{r^2}}}$),would Gauss' law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that the electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant ($= 80$) than say, mica ($= 6$).

Answer 1

Gauss’s law states that the total electric flux due to closed surface is equal to the $\dfrac{1}{{{\varepsilon _0}}}$ times charge enclosed by the surface, Coulomb’s law for two charges states that the total force acting on a point charge due to other charge proportional to the product of their charges and inversely proportional to the square of their distance . Work done on a body is given by the dot product of the force and displacement of the body.

Gauss’s law is given by, $\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{q}{{{\varepsilon _0}}}$where, $\vec E$ is the electric field, $S$ is the closed surface placed in the electric field. Direction of $S$ is always normal to it. Coulomb’s force is given by, $\overrightarrow {{F_x}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}}{\kern 1pt} \widehat x$ Where, $Q$ is a charge separated by distance $x$ from another charge $q$ then Force on each charge by the other charge is ${F_x}$ , ${\varepsilon _0}$ is the permittivity of vacuum or air.

Complete step by step answer:
(a) We have here two large conducting spheres with charge ${Q_1}$​ and ${Q_2}$​ placed closed. Now, we know Coulomb’s law states that the force acting on a point charge due to other charge $\overrightarrow {{F_x}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}}{\kern 1pt} \widehat x$ Where, $Q$ is a charge separated by distance $x$ from another charge $q$ then Force on each charge by the other charge is ${F_x}$ , ${\varepsilon _0}$ is the permittivity of vacuum or air.

Now, we know if we have two spheres containing charges ${Q_1}$ and ${Q_2}$ they will act as a point charge when they are separated by a large distance. But, when they are close then the charge distribution on them cannot be taken as point charges. Hence, force acting on the spheres will not be equal to $\dfrac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$.

(b) We know, Gauss’s law states that, that the total electric flux due to closed surface is equal to the $\dfrac{1}{{{\varepsilon _0}}}$ times charge enclosed by the surface. Mathematically, , $\phi = \oint\limits_s {\vec E} .d\vec S = \dfrac{q}{{{\varepsilon _0}}}$ where, $\vec E$ is the electric field, $S$ is the closed surface placed in the electric field. Direction of $S$ is always normal to it.

Now, we know here, $\vec E$ varies as $\dfrac{1}{{{r^2}}}$ when coulomb’s law involves $\dfrac{1}{{{r^2}}}$ and $d\vec S$ varies as ${r^2}$. So, $\phi$ is independent of $r$. Now, if coulomb’s law involves $\dfrac{1}{{{r^3}}}$ then $\phi$ will vary as $\propto \dfrac{1}{{{r^3}}} \cdot {r^2} \propto r$. Hence, Gauss’s law will not be true.

(c) We know that the direction of acceleration is always in the direction of force. So, for an electrostatic field the direction of the field always gives the direction of acceleration at that point. Now, here we have a small test charge, which is released from rest at a point in an electrostatic field. So, it will travel along the field line passing through that point.

(d) We know that, Work done on a body is given by the dot product of the force and displacement of the body. So, when the electron completes an orbit, either circular or elliptical, the displacement is always zero, so the work done by the field of the nucleus will be zero.

(e) We know, the electric field is discontinuous across the surface of a charged conductor, but the electric potential is always continuous. So, the electric potential is continuous across the surface.

(f) If only the conductor is placed then we can think that the conductor is one plate of a parallel plate capacitor while the other plate is placed at infinity.So, the capacitance of a single conductor means that the capacitance of a parallel plate capacitor with one plate as the conductor, while the other conductor is placed at infinity.

(g) We know that the dipole moment of two opposite charges is given by, $p = ql$ where $l$ is the distance between them. Now, dielectric constant depends on the dipole moment per unit volume of the substance. So, water has a large dipole moment per unit volume due to its asymmetric structure while mica is a symmetrical crystal. So, water has a large dielectric constant compared to mica.

Note: Tangent drawn at a point on the field line gives the direction of acceleration at that point.
$\bullet$ Electric field inside a conductor is always zero, that is why it is discontinuous across the surface but potential is always constant since, $\vec E = - \nabla V$.
$\bullet$ The Dipole moment per unit volume of a material is called the Polarization of the material. It is related to the dielectric constant as, $\vec P = {\varepsilon _0}\chi \vec E = {\varepsilon _0}(K - 1)\vec E$. Where, $\chi$ is the susceptibility and $K$ is the dielectric constant of the material.