Question
Derive expression for electric field intensity due to a point charge in a material medium.

Answer

i. Consider a point charge q placed at point O in a medium of dielectric constant K as shown in figureImage
ii. Consider the point P in the electric field of point charge at distance r from q. A test charge q0 placed at the point P will experience a force which is given by the Coulomb’s law,
$\vec{F}=\frac{1}{4 \pi \varepsilon_0 K} \frac{ qq _0}{ r ^2} \hat{ r }$
where $\hat{r}$ is the unit vector in the direction of force i.e., along $O P$.
iii. By the definition of electric field intensity,
$\vec{F}=\frac{F}{ q _0}=\frac{1}{4 \pi \varepsilon_0 K} \frac{ q }{ r ^2} \hat{ r }$
The direction of $\vec{E}$ will be along OP when $q$ is positive and along PO when q is negative.
iv. The magnitude of electric field intensity in a medium is given by, $E=\frac{1}{4 \pi \varepsilon_0 K} \frac{q}{r^2}$
v. For air or vacuum, K = 1 then
$E =\frac{1}{4 \pi \varepsilon_0} \frac{ q }{ r ^2}$

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