Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

Demonstrate that the potential of the field generated by a dipole, with electric moment $\vec{P}$ , may be represented as $φ = \frac{\vec{p} \cdot \vec{r}}{4 π ε_{0} r^{3}}$ , where $\vec{r}$ is the radius vector. Using this expression, find the magnitude of the electric field strength vector as a function of $r$ and $θ$ .

Accepted Answer

Assume that there are two charges, $+ q$ and $- q$ , separated by a distance $l$ .

Then, the electric potential at a point, at a distance $r > > l$ from this dipole, is

$φ (r) = \frac{+ q}{4 π ε_{0} r_{+}} + \frac{- q}{4 π ε_{0} r_{-}} = \frac{q}{4 π ε_{0}} (\frac{r_{-} - r_{+}}{r_{+} r_{-}}) . . . (1)$

where, $r_{-} =$ Distance of point $S$ from charge $- q$ ,

$r_{+} =$ Distance of point $S$ from charge $+ q$ .

But, $r_{-} - r_{+} = l cosθ$ and $r_{+} r_{-} ≅ r^{2} . . . (2)$

From equations $(1)$ and $(2)$ ,

$φ (r) = \frac{q l cosθ}{4 π ε_{0} r^{2}} = \frac{p cosθ}{4 π ε_{0} r^{2}} φ = \frac{\vec{p} \cdot \vec{r}}{4 π ε_{0} r^{3}}$

where, $p$ is the magnitude of electric moment vector.

Now, $E_{r} = - \frac{\partial φ}{\partial r} = \frac{2 p cosθ}{4 π ε_{0} r^{3}}$

and $E_{θ} = - \frac{\partial φ}{r \partial θ} = \frac{p sinθ}{4 π ε_{0} r^{3}}$

So, $E = \sqrt{E_{r}^{2} + E_{0}^{2}} = \frac{p}{4 π ε_{0} r^{3}} \sqrt{4 \cos^{2} θ + \sin^{2} θ}$

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

Demonstrate that the potential of the field generated by a dipole, with electric moment P→, may be represented as φ=p→·r→4πε0r3, where r→ is the radius vector. Using this expression, find the magnitude of the electric field strength vector as a function of r and θ.

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !