Question

Two charges – q and + q are located at points (0, 0, – a) and (0, 0, a) respectively.
(a) What 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 x-axis?

Solution

Two charges -q and +q are located at points A and B as shown in the above fig.

i) Electrostatic potential at point (0,0,z) :

V = \frac{1}{4 π ε_{\circ}} \frac{- q}{A P} + \frac{1}{4 π ε_{\circ}} \frac{+ q}{B P} = \frac{q}{4 π ε_{\circ}} (\frac{1}{B P} - \frac{1}{A P}) where, AP = (z + a) and BP = (z - a) = \frac{q}{4 {πε}_{\circ}} (\frac{1}{z - a} - \frac{1}{z + a}) = \frac{q}{4 π ε_{\circ}} (\frac{z + a - (z - a)}{z^{2} - a^{2}}) = \frac{1}{4 π ε_{\circ}} \frac{q . 2 a}{z^{2} - a^{2}} = \frac{1}{4 π ε_{\circ}} . \frac{p}{z^{2} - a^{2}} [p = q . 2 a is the dipole moment of the given charges .]

Electrostatic potential at point (x,y,0):

V = \frac{1}{4 {πε}_{\circ}} . \frac{- q}{AQ} + \frac{1}{4 {πε}_{\circ}} . \frac{+ q}{BQ} = \frac{q}{4 {πε}_{\circ}} (\frac{1}{BQ} - \frac{1}{AQ}) where, AQ = \sqrt{(x - 0)^{2} + (y - 0)^{2} + (0 - (- a)^{2}} = \sqrt{x^{2} + y^{2} + a^{2}} BQ = \sqrt{(x - 0)^{2} + (y - 0)^{2} + (0 - a)^{2}} = \sqrt{x^{2} + y^{2} + a^{2}}

Since, AQ=BQ
We have, electric potential at (x,y,0)=0.

ii) Electrostatic potential is obtained as

V = \frac{1}{4 π ε_{\circ}} . \frac{p}{r^{2} - a^{2}}

Under the given condition when,

\frac{r}{a} > > 1 \Rightarrow a < < r

and, the above equation reduces to

V = \frac{1}{4 {πε}_{\circ}} . \frac{p}{r^{2}}

.

Therefore,

V \propto \frac{1}{r^{2}}

.

iii) The amount of work done to move a small test charge from point (5,0,0) to (-7,0,0) is the potential difference at these points.

∴ Potential at (5, 0, 0), V_{1} = \frac{1}{4 {πε}_{\circ}} [\frac{q}{\sqrt{(5 - 0)^{2} - a^{2}}} - \frac{q}{\sqrt{(5 - 0)^{2} - (- a)^{2}}}] = 0 Potential at (- 7, 0, 0), V_{2} = \frac{1}{4 {πε}_{\circ}} [\frac{q}{\sqrt{(- 7 - 0)^{2} - a^{2}}} - \frac{q}{\sqrt{(- 7 - 0)^{2} - (- a)^{2}}}] = 0

Work done, W= q_o(V₂-V₁) = 0

The answer will not change if the same test charge is moved between the same points along some other axis because, work done does not depend upon the path followed.

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