2.22 Figure 2.32 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

 

2.22 Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge -2q placed at Y

Charge +q placed at point Z.

XY = YZ = a

YP = r

PX = r + a

PZ = r – a

Electrostatic potential caused by the system of three charges at point P is given by,

V = $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{XP}$ - $\frac{2q}{YP}$ + $\frac{q}{ZP}$ ]

= $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{r+a}$ - $\frac{2q}{r}$ + $\frac{q}{r-a}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r\left(r-a\right)-2\left(r-a\right)\left(r+a\right)+r(r+a)}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{2a^2}{r(r^2-a^2)}$ ]

= $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3(1-\frac{a^2}{r^2})}$

Since $\frac{r}{a}?1,$ therefore $\frac{a}{r}?1$ . So $\frac{a^2}{r^2}$ is taken as negligible.

Therefore V = $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3}$

It can be inferred t

...more

2.22 Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge -2q placed at Y

Charge +q placed at point Z.

XY = YZ = a

YP = r

PX = r + a

PZ = r – a

Electrostatic potential caused by the system of three charges at point P is given by,

V = $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{XP}$ - $\frac{2q}{YP}$ + $\frac{q}{ZP}$ ]

= $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{r+a}$ - $\frac{2q}{r}$ + $\frac{q}{r-a}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r\left(r-a\right)-2\left(r-a\right)\left(r+a\right)+r(r+a)}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{2a^2}{r(r^2-a^2)}$ ]

= $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3(1-\frac{a^2}{r^2})}$

Since $\frac{r}{a}?1,$ therefore $\frac{a}{r}?1$ . So $\frac{a^2}{r^2}$ is taken as negligible.

Therefore V = $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3}$

It can be inferred that V $\propto \frac{1}{r^3}$ . It is known that for a dipole V $\propto \frac{1}{r^2}$ and for monopole V $\propto \frac{1}{r}$

less

<p><strong>2.22 </strong>Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728972351php9OXCBs_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728972351php9OXCBs.jpeg" alt="" width="350" height="54"></picture></div></div><p>It can be considered that the system of the electric quadrupole has three charges.</p><p>Charge +q placed at point X</p><p>Charge -2q placed at Y</p><p>Charge +q placed at point Z.</p><p>XY = YZ = a</p><p>YP = r</p><p>PX = r + a</p><p>PZ = r – a</p><p>Electrostatic potential caused by the system of three charges at point P is given by,</p><p>V =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mo>[</mo><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mi>X</mi><mi>P</mi></mrow></mrow></mfrac></math></span>&nbsp;-&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mi>q</mi></mrow></mrow><mrow><mrow><mi>Y</mi><mi>P</mi></mrow></mrow></mfrac></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mi>Z</mi><mi>P</mi></mrow></mrow></mfrac></math></span>&nbsp;]</p><p>=&nbsp;<span title="Click to copy mathml"><math><mi></mi><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mo>[</mo><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mi>r</mi><mo>+</mo><mi>a</mi></mrow></mrow></mfrac></math></span>&nbsp;-&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mi>q</mi></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></mfrac></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mi>r</mi><mo>-</mo><mi>a</mi></mrow></mrow></mfrac></math></span>&nbsp;]</p><p>=&nbsp;<span title="Click to copy mathml"><math><mi></mi><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mo>[</mo><mfrac><mrow><mrow><mi>r</mi><mfenced separators="|"><mrow><mrow><mi>r</mi><mo>-</mo><mi>a</mi></mrow></mrow></mfenced><mo>-</mo><mn>2</mn><mfenced separators="|"><mrow><mrow><mi>r</mi><mo>-</mo><mi>a</mi></mrow></mrow></mfenced><mfenced separators="|"><mrow><mrow><mi>r</mi><mo>+</mo><mi>a</mi></mrow></mrow></mfenced><mo>+</mo><mi>r</mi><mo>(</mo><mi>r</mi><mo>+</mo><mi>a</mi><mo>)</mo></mrow></mrow><mrow><mrow><mi>r</mi><mfenced separators="|"><mrow><mrow><mi>r</mi><mo>-</mo><mi>a</mi></mrow></mrow></mfenced><mo>(</mo><mi>r</mi><mo>+</mo><mi>a</mi><mo>)</mo></mrow></mrow></mfrac></math></span>&nbsp;]</p><p>=&nbsp;<span title="Click to copy mathml"><math><mi></mi><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mo>[</mo><mfrac><mrow><mrow><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mi>r</mi><mi>a</mi><mo>-</mo><mn>2</mn><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><mn>2</mn><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><mi>r</mi><mi>a</mi></mrow></mrow><mrow><mrow><mi>r</mi><mfenced separators="|"><mrow><mrow><mi>r</mi><mo>-</mo><mi>a</mi></mrow></mrow></mfenced><mo>(</mo><mi>r</mi><mo>+</mo><mi>a</mi><mo>)</mo></mrow></mrow></mfrac></math></span>&nbsp;]</p><p>=&nbsp;<span title="Click to copy mathml"><math><mi></mi><mfrac><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mo>[</mo><mfrac><mrow><mrow><mn>2</mn><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mi>r</mi><mo>(</mo><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mi></mi><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>)</mo></mrow></mrow></mfrac></math></span>&nbsp;]</p><p>=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mi>q</mi><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup><mo>(</mo><mn>1</mn><mo>-</mo><mfrac><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>)</mo></mrow></mrow></mfrac></math></span></p><p>Since&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mi>a</mi></mrow></mrow></mfrac><mi></mi><mo>?</mo><mn>1</mn><mo>,</mo><mi></mi></math></span>&nbsp;therefore&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></mfrac><mo>?</mo><mn>1</mn></math></span>&nbsp;. So&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;is taken as negligible.</p><p>Therefore V =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mi>q</mi><msup><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span></p><p>It can be inferred that V&nbsp;<span title="Click to copy mathml"><math><mo>∝</mo><mi></mi><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;. It is known that for a dipole V&nbsp;<span title="Click to copy mathml"><math><mo>∝</mo><mi></mi><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;and for monopole V&nbsp;<span title="Click to copy mathml"><math><mo>∝</mo><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mi>r</mi></mrow></mrow></mfrac></math></span></p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment