Electrostatics IIT JEE Study Material

According to the Coulomb’s law of electrostatics, the magnitude of the repulsive or attractive electrostatic force (F)between the 2 point charges \(\mathbf{q_{1}}\) and \(\mathbf{q_{2}}\) is directly proportional to the product of the magnitudes of charges i.e. \(\mathbf{F\;\propto \; q_{1}\;.\;q_{2}}\) and inversely proportional to the square of the distance (r) between \(\mathbf{q_{1}}\) and \(\mathbf{q_{2}}\) i.e. \(\mathbf{F\;\propto \; \frac{1}{r^{2}}}\). This electrostatic force acts along the straight line joining them. This electrostatic force is repulsive in nature when both the charges \(\mathbf{q_{1}}\) and \(\mathbf{q_{2}}\) have the same sign. The Electrostatic force between them is attractive if both the charges \(\mathbf{q_{1}}\) and \(\mathbf{q_{2}}\) have different signs. This Law is applicable only for the static and the point charges.

Coulomb’s Law: \(\mathbf{F\;=\;\frac{1}{4\;\pi \;\epsilon _{0} \;\epsilon _{r}}\; \times \;\frac{q_{1}\;q_{2}}{r^{2}}}\)

And, In the vector Form:

\(\mathbf{\overrightarrow{F}\;=\;\frac{1}{4\;\pi \;\epsilon _{0} \;\epsilon _{r}}\; \times \;\frac{q_{1}\;q_{2}}{r^{3}}\;\overrightarrow{r}}\)

Where,

\(\mathbf{\epsilon _{0}}\) = Permittivity of the free space = \(\mathbf{8.85\; \times \;10^{-12}\;N^{-1}\;m^{-2}\;c^{2}}\)

\(\mathbf{\epsilon _{r}}\) = Relative permittivity of the medium [For Air \(\mathbf{\epsilon _{r}\;=\;1}\)]

\(\mathbf{\epsilon _{0}}\) \(\mathbf{\epsilon _{r}}\) = Absolute permittivity of the medium

r = Distance between charge \(\mathbf{q_{1}}\) and \(\mathbf{q_{2}}\)

Electric Field Due to the Point Charge (Vector Form):

\(\mathbf{\overrightarrow{E}\;=\;\frac{1}{4\;\\pi \;\epsilon _{0}} \; \times \;\frac{q}{r^{2}}\;\;\hat{r}=\frac{1}{4\;\pi \;\epsilon _{0}} \; \times \;\frac{q}{r^{3}}\;\;\overrightarrow{r}}\)

Electric Field Due to the continuous charge distribution:

\(\mathbf{\overrightarrow{E}\;=\;\frac{1}{4\;\pi \;\epsilon _{0}}\;int \;\frac{dq}{r^{2}}\;\hat{r}\;=\;\int d\overrightarrow{E}}\)

Where, \(\mathbf{ d\overrightarrow{E}}\) = Electric Field Due to an elementary charge

Electric Field Due to the Infinite line of Charge:

\(\mathbf{ \overrightarrow{E}\;=\;\frac{2\;k\;\lambda }{r}}\)

Where \(\mathbf{\lambda }\) = linear charge density

Electric Field Due to the Uniformly Charged ring:

\(\mathbf{ E_{centre}\;=\;0\;\;and\;\;E_{axis}\;=\;\frac{k\;Q\;x }{\left ( x^{2}\;+\;R^{2} \right )^{\frac{3}{2}}}}\)

Potential Due to a point charge:

\(\mathbf{V\;=\;\frac{Q}{4\;\pi \;\epsilon _{0}\;r} }\)

Potential Due to several charges:

\(\mathbf{V\;=\;\frac{q_{1}}{4\;\pi \;\epsilon _{0}\;r_{1}} \;+\;\frac{q_{2}}{4\;\pi \;\epsilon _{0}\;r_{2}} \;+\;\frac{q_{3}}{4\;\pi \;\epsilon _{0}\;r_{3}}\;+\;.\;.\;.\;.\;}\)

Potential Due to the spherical shell or solid conducting sphere:

\(\mathbf{V_{in}\;=\;\frac{Q}{4\;\pi \;\epsilon _{0}\;R}\;[r\leq R]\;\;and\;\;V_{out}\;=\;\frac{Q}{4\;\pi \;\epsilon _{0}\;r}\;\;[r\geq R]}\)

Potential Due to the non-conducting uniformly charged solid sphere:

\(\mathbf{V_{in}\;=\;\frac{Q\left ( 3R^{2}\;-\;r^{2} \right )}{8\;\pi \;\epsilon _{0}\;R}\;[r\leq R]\;\;and\;\;V_{out}\;=\;\frac{Q}{4\;\pi \;\epsilon _{0}\;r}\;\;[r\geq R]}\)

Potential Due to the continuous charge distribution:

\(\mathbf{V\;=\;\frac{1}{4\;\pi \;\epsilon _{0}}\;\;\int \;\frac{dq}{r}}\)

Electric Flux:

Electric flux in a uniform electric field (E) is given by:

\(\mathbf{\Phi =\overrightarrow{E}\;.\;\overrightarrow{A}\;=\;E\;A\;cos\;\theta }\)

Where, \(\mathbf{ heta }\) = Angle between Area vector \(\mathbf{\overrightarrow{A} }\) and \(\mathbf{\overrightarrow{E} }\)

In case of the non-uniform Electric Field,

\(\mathbf{\Phi \;=\;\int \overrightarrow{E}\;.\;d\overrightarrow{A}}\)

Gauss’s Law: The net flux emerging out of a closed surface is \(\mathbf{\frac{q}{\epsilon _{0}}}\)

i.e. \(\mathbf{\Phi \;=\;\oint \overrightarrow{E}\;d\overrightarrow{A}\;=\;\frac{q}{\epsilon _{0}}}\)

Where q = net charge enclosed by the closed surface

Note: The value of \(\mathbf{\Phi}\) does not depend on the size and shape of the closed surface and the charges located outside the closed surface.

Electric Dipole:

Electrostatics for IIT JEE

Case 1: On the Axis

If r << a then, \(\mathbf{\overrightarrow{E}\;=\;\frac{\overrightarrow{p}\;r}{2\;\pi \;\epsilon _{0}\;\left [ r^{2} \;-\;\left ( \frac{a^{2}}{4} \right ) \right ]^{2}}\;\;i.e\;approx \;=\;\frac{\overrightarrow{p}}{2\;\pi \;\epsilon _{0}\;r^{3}}}\)

Case 2: On the Equatorial

\(\mathbf{\overrightarrow{E}\;=\;\frac{\overrightarrow{p}}{4\;\pi \;\epsilon _{0}\;\left [ r^{2} \;+\;\left ( \frac{a^{2}}{4} \right ) \right ]^{\frac{3}{2}}}\;\;i.e\;approx \;=\;\frac{-\;\overrightarrow{p}}{4\;\pi \;\epsilon _{0}\;r^{3}}}\)

Where,

r = the distance of the point from the center of the dipole

\(\mathbf{\overrightarrow{p}\;=\;q\;\overrightarrow{a}}\) = Dipole Moment

Force on a dipole when placed in a non-uniform electric field:

\(\mathbf{F\;=\;-\;\frac{d}{dx}\;\left ( -\;\overrightarrow{P}\;.\;\overrightarrow{E} \right )\;\hat{i}\;=\;\overrightarrow{P}\;\frac{d\overrightarrow{E}}{dx}\;\;\hat{i}}\)


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