Search any question & find its solution
Question:
Answered & Verified by Expert
Two charges $-q$ and $+q$ are located at points $(0,0,-$ a) and $(\mathbf{0}, \mathbf{0}$, a), respectively.
(a) Why is the electrostatic potential at the points $(0,0, z)$ and $(x, y, 0)$ ?
(b) Obtain the dependence of potential on the distance $r$ of a point from the origin when $r / a>1$.
(c) How much work is done in moving a small test charge from the point $(5,0,0)$ to $(-7,0,0)$ along the $x$-axis? Does the answer change if the path of the test charge between the same points is not along the $\mathrm{x}$-axis?
(a) Why is the electrostatic potential at the points $(0,0, z)$ and $(x, y, 0)$ ?
(b) Obtain the dependence of potential on the distance $r$ of a point from the origin when $r / a>1$.
(c) How much work is done in moving a small test charge from the point $(5,0,0)$ to $(-7,0,0)$ along the $x$-axis? Does the answer change if the path of the test charge between the same points is not along the $\mathrm{x}$-axis?
Solution:
1590 Upvotes
Verified Answer
(a) Distance between points $(0,0,-a)$ and $(0,0, z)=z+a$
Distance between points $(0,0$, a) and $(0,0, z)=z-a$
Potential due to charge $-\mathrm{q}$ at point $(0,0, \mathrm{z})=\frac{1}{4 \pi \varepsilon_0} \frac{-\mathrm{q}}{\mathrm{z}+\mathrm{a}}$
Potential due to charge $+\mathrm{q}$ at point $(0,0, z)=\frac{1}{4 \pi \varepsilon_0} \frac{q}{z-a}$
Total potential at point $(0,0, z)=-\frac{q}{4 \pi \varepsilon_0(z+a)}+\frac{q}{4 \pi \varepsilon_0(z-a)}$ $=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{(z-a)}-\frac{1}{(z+a)}\right]$ $=\frac{\mathrm{q}}{4 \pi \varepsilon_0} \times \frac{2 \mathrm{a}}{\mathrm{z}^2-\mathrm{a}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{p}}{\mathrm{z}^2-\mathrm{a}^2} \quad$ where $\mathrm{p}$ $=\mathrm{q} \cdot 2 \mathrm{a}=$ dipole moment of charge system. Distance between points $(0,0,-a)$ and $(x, y, 0)=\sqrt{x^2+y^2+a^2}$
Distance between points $(0,0$, a) and $(x, y, 0)=\sqrt{x^2+y^2+a^2}$
Since the two distances are equal and charges are equal and opposite, total potential at point $(x, y$, $0)=$ zero.
(b) The dependence on $\mathrm{r}$ is $1 / \mathrm{r}^2$ type
(c) Zero; no, because work done by electrostatic field between two points is independent of the path connecting the two points.
Distance between points $(0,0$, a) and $(0,0, z)=z-a$
Potential due to charge $-\mathrm{q}$ at point $(0,0, \mathrm{z})=\frac{1}{4 \pi \varepsilon_0} \frac{-\mathrm{q}}{\mathrm{z}+\mathrm{a}}$
Potential due to charge $+\mathrm{q}$ at point $(0,0, z)=\frac{1}{4 \pi \varepsilon_0} \frac{q}{z-a}$
Total potential at point $(0,0, z)=-\frac{q}{4 \pi \varepsilon_0(z+a)}+\frac{q}{4 \pi \varepsilon_0(z-a)}$ $=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{(z-a)}-\frac{1}{(z+a)}\right]$ $=\frac{\mathrm{q}}{4 \pi \varepsilon_0} \times \frac{2 \mathrm{a}}{\mathrm{z}^2-\mathrm{a}^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{p}}{\mathrm{z}^2-\mathrm{a}^2} \quad$ where $\mathrm{p}$ $=\mathrm{q} \cdot 2 \mathrm{a}=$ dipole moment of charge system. Distance between points $(0,0,-a)$ and $(x, y, 0)=\sqrt{x^2+y^2+a^2}$
Distance between points $(0,0$, a) and $(x, y, 0)=\sqrt{x^2+y^2+a^2}$
Since the two distances are equal and charges are equal and opposite, total potential at point $(x, y$, $0)=$ zero.
(b) The dependence on $\mathrm{r}$ is $1 / \mathrm{r}^2$ type
(c) Zero; no, because work done by electrostatic field between two points is independent of the path connecting the two points.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.