Name: Electric Potential Due to Discrete Charges
Uploaded: 2019-07-19
Duration: 3 min 25 s
Description: How do we calculate the net potential at a point due to a system of charges? Hint: Electric potential is a scalar quantity! Watch the video to learn the calc...

Question

Derive an expression for the electric potential due to a point charge.

A positive charge of magnitude $6 μC$ is placed in a dielectric medium of dielectric constant $2$ . What is the electric potential at a point in the dielectric medium at a distance of $10 cm$ from the charge ?

Accepted Answer

Let $+ q$ be an isolated point charge situated in air at $O$ . $P$ is a point at a distance $r$ from $+ q$ .

Consider two points $A$ and $B$ at distances $x$ and $x + d x$ from the point $O$ .

The potential difference between $A$ and $B$ is, $d V = - E d x$ .................... $(1)$

The electric field strength at $A$ due to the point charge is, $E = \frac{1}{4 π ε_{\circ}} \frac{q}{x^{2}}$ .................... $(2)$

By substituting equation $(2)$ in equation $(1)$ we get: $d V = - \frac{1}{4 π ε_{\circ}} \frac{q}{x^{2}} d x$ .................... $(3)$

Here, the negative sign indicates that the work is done against the electric force.

The electric potential at the point $P$ due to the charge $+ q$ is the total work done in moving a unit positive charge from infinity to that point.

$V = \int_{\infty}^{r} - \frac{1}{4 π ε_{\circ}} \frac{q}{x^{2}} d x$

$V = \frac{q}{4 π ε_{\circ}} (\frac{1}{r} - \frac{1}{\infty})$

$V = \frac{1}{4 π ε_{\circ}} \frac{q}{r}$

Hence, the electric potential due to a point charge is $\frac{1}{4 π ε_{\circ}} \frac{q}{r}$ .

Now a point charge of $q = 6 μC$ is placed in a dielectric medium with dielectric constant $2$ . Then the electric potential at the given point due to this point charge: $V' = \frac{1}{4 π ε_{\circ} ε_{r}} \frac{q}{r}$

$V' = \frac{9 \times 10^{9} N m^{2} C^{- 2}}{2} \frac{6 \times 10^{- 6} C}{10^{- 1} m} = 27 \times 10^{4} V$

Hence, the electric potential at the given point is $27 \times 10^{4} V$ .

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