- #1
null void
- 102
- 1
Homework Statement
- A point charge is placed at the origin of the medium.
- The relative permittivity of the medium, [itex]\varepsilon_r = a / r[/itex], a is a constant, r is the radius from origin to any point around the charge.
- Objective of this question is to find the expression for voltage.at any point around Q
Homework Equations
The Attempt at a Solution
[itex]\vec E = \frac{Q}{4\pi \varepsilon_0 \varepsilon_r r^2} \vec r[/itex]
[itex] = \frac{Q}{4\pi \varepsilon_0 (\frac{a}{r}) r^2} \vec r [/itex]
[itex] = \frac{Q}{4\pi \varepsilon_0 a r }\vec r [/itex]
[itex]V = - \int_\infty^r{\vec E \cdot d \vec l}[/itex]
[itex] = - \int_\infty^r{ \frac{Q}{4\pi \varepsilon_0 a r } dr}[/itex]
[itex] = - \frac{Q}{4\pi \varepsilon_0 a}[ln r - ln \infty] [/itex]
[itex] = \frac{Q}{4\pi \varepsilon_0 a}[\ln \frac{\infty}{r}] [/itex]
In the integration i use [itex]\infty[/itex] as lower boundary to assume the infinity is at 0 voltage.
But the expression seems to be wrong, anyone know where did I messed up?