2.14 Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) At the mid-point of the line joining the two charges, and

(b) At a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

2.14

Two charges are at point A and B. O is the midpoint of line joining A & B

Magnitude of charge at point A, $q_1$ = 1.5 $\mu C$ = 1.5 $\times {10}^{-6}C$ and

at point B, $q_2$ = 2.5 $\mu C=2.5\times {10}^{-6}C$

Distance between A & B, d = 30 cm = 0.3 m

Let $V_1$ and $E_1$ be the electric potential and electric field respectively at O.

$V_1=$ Potential due to charge at A + Potential due to charge at B

$=\frac{q_1}{4\pi {\epsilon }_0\frac{d}{2}}$ + $\frac{q_2}{4\pi {\epsilon }_0\frac{d}{2}}$ = $\frac{1}{4\pi {\epsilon }_0\frac{d}{2}}$ ( $q_1+q_2)$

${\epsilon }_0$ = permittivity of free space = 8.854 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

$V_1=$ $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times \frac{0.3}{2}}$ (1.5 $\times {10}^{-6}+2.5\times {10}^{-6})$ = 2.4 $\times {10}^5$ V

$E_1=$ Electric field due to $q_2$ - Electric field due to $q_1$

$=$ $\frac{q_2}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$ $-$ $\frac{q_1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$

$=$ $\frac{1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}(q_2-q_1)$ = $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times ({\frac{0.3}{2})}^2}(2.5\times {10}^{-6}-1.5\mathrm{}\times {10}^{-6})$

= 4

...more

2.14

Two charges are at point A and B. O is the midpoint of line joining A & B

Magnitude of charge at point A, $q_1$ = 1.5 $\mu C$ = 1.5 $\times {10}^{-6}C$ and

at point B, $q_2$ = 2.5 $\mu C=2.5\times {10}^{-6}C$

Distance between A & B, d = 30 cm = 0.3 m

Let $V_1$ and $E_1$ be the electric potential and electric field respectively at O.

$V_1=$ Potential due to charge at A + Potential due to charge at B

$=\frac{q_1}{4\pi {\epsilon }_0\frac{d}{2}}$ + $\frac{q_2}{4\pi {\epsilon }_0\frac{d}{2}}$ = $\frac{1}{4\pi {\epsilon }_0\frac{d}{2}}$ ( $q_1+q_2)$

${\epsilon }_0$ = permittivity of free space = 8.854 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

$V_1=$ $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times \frac{0.3}{2}}$ (1.5 $\times {10}^{-6}+2.5\times {10}^{-6})$ = 2.4 $\times {10}^5$ V

$E_1=$ Electric field due to $q_2$ - Electric field due to $q_1$

$=$ $\frac{q_2}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$ $-$ $\frac{q_1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$

$=$ $\frac{1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}(q_2-q_1)$ = $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times ({\frac{0.3}{2})}^2}(2.5\times {10}^{-6}-1.5\mathrm{}\times {10}^{-6})$

= 4.0 $\times {10}^5$ V/m

Therefore, the potential at mid-point is 2.4 $\times {10}^5$ V and the electric field is

4.0 $\times {10}^5$ V/m and it is directed from larger charge to smaller charge.

 

Let us consider a point Z such that normal distance OZ = 10 cm = 0.1 m

Let $V_2$ and $E_2$ be the electric potential and electric field respectively at Z.

From the $?$ ABZ, we can write,

BZ = AZ = $\sqrt{{0.15}^2+{0.1}^2}$ = 0.18 m

$V_2=$ Potential due to charge at A + Potential due to charge at B

$=\frac{q_1}{4\pi {\epsilon }_{0}AZ}$ + $\frac{q_2}{4\pi {\epsilon }_{0}BZ}$ = $\frac{1}{4\pi {\epsilon }_{0}0.18}$ ( $q_1+q_2)$

${\epsilon }_0$ = permittivity of free space = 8.854 $\times {10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

$V_2=$ $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}\times 0.18}$ (1.5 $\times {10}^{-6}+2.5\times {10}^{-6})$ = 2.0 $\times {10}^5$ V

Electric field due to charge $q_{1}atZ,E_A$

$E_A=\frac{q_1}{4\pi {\epsilon }_0({AZ)}^2}$ = $\frac{1.5\mathrm{}\times {10}^{-6}}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}\times ({0.18)}^2}$ = 0.416 $\times {10}^6$ V/m

Electric field due to charge $q_{2}atZ,E_B$

$E_B=\frac{q_2}{4\pi {\epsilon }_0({BZ)}^2}$ = $\frac{2.5\mathrm{}\times {10}^{-6}}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}\times ({0.18)}^2}$ = 0.693 $\times {10}^6$ V/m

The resultant intensity at Z

$E_2$ = $\sqrt{{E_A}^2+{E_B}^2+2E_A{E}_{B}cos2\theta }$ where 2 $\theta$ is the $\angle AZB$

From the figure, we get $\cos ?\theta$ = $\frac{ZO}{ZB}$ = $\frac{0.1}{0.18}$

Hence $\theta$ = 56.25 $°$

Therefore, $E_2$ = $E_2=\sqrt{({0.416\mathrm{}\times {10}^6)}^2+{(0.693\mathrm{}\times {10}^6)}^2+2{\left(0.416\mathrm{}\times {10}^6\right)\times (0.693\mathrm{}\times {10}^6)}_{}cos2\times 56.25\mathrm{}°}$ = 6.57 $\times {10}^5$ V/m

less

<p><strong>2.14</strong></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728972222phprnTZi1_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728972222phprnTZi1.jpeg" alt="" width="399" height="97"></picture></div><div><p>Two charges are at point A and B. O is the midpoint of line joining A &amp; B</p><p>Magnitude of charge at point A,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;= 1.5&nbsp;<span title="Click to copy mathml"><math><mi>μ</mi><mi>C</mi></math></span>&nbsp;= 1.5&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mi>C</mi><mi></mi></math></span>&nbsp;and</p><p>at point B,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;= 2.5&nbsp;<span title="Click to copy mathml"><math><mi>μ</mi><mi>C</mi><mo>=</mo><mn>2.5</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mi>C</mi></math></span></p><p>Distance between A &amp; B, d = 30 cm = 0.3 m</p><p>Let&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn><mi></mi></mrow></mrow></msub></math></span>&nbsp;and&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;be the electric potential and electric field respectively at O.</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn><mi></mi></mrow></mrow></msub><mo>=</mo></math></span>&nbsp;Potential due to charge at A + Potential due to charge at B</p><p><span title="Click to copy mathml"><math><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></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><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfrac></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></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><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfrac></math></span>&nbsp;=&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><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfrac></math></span>&nbsp;(&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>+</mo><mi></mi><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>)</mo></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math></span>&nbsp;= permittivity of free space = 8.854&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup></math></span>&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><msup><mrow><mrow><mi>N</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math></span>&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn><mi></mi></mrow></mrow></msub><mo>=</mo></math></span>&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><mo>×</mo><mn>8.854</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup><mo>×</mo><mfrac><mrow><mrow><mn>0.3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></mrow></mrow></mfrac></math></span>&nbsp;(1.5&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>+</mo><mi></mi><mn>2.5</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>)</mo></math></span>&nbsp;= 2.4&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;V</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>=</mo><mi></mi></math></span>&nbsp;Electric field due to&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;- Electric field due to&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span></p><p><span title="Click to copy mathml"><math><mo>=</mo><mi></mi></math></span>&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></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><mo>(</mo><msup><mrow><mrow><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;<span title="Click to copy mathml"><math><mo>-</mo></math></span>&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></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><mo>(</mo><msup><mrow><mrow><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span></p><p><span title="Click to copy mathml"><math><mo>=</mo></math></span>&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><mo>(</mo><msup><mrow><mrow><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>(</mo><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>-</mo><mi></mi><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>)</mo></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><mo>×</mo><mn>8.854</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup><mo>×</mo><mo>(</mo><msup><mrow><mrow><mfrac><mrow><mrow><mn>0.3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac><mo>(</mo><mn>2.5</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>-</mo><mi></mi><mn>1.5</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>)</mo></math></span></p><p>= 4.0&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;V/m</p><p>Therefore, the potential at mid-point is 2.4&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;V and the electric field is</p><p>4.0&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;V/m and it is directed from larger charge to smaller charge.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728972246phpju2u7D_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728972246phpju2u7D.jpeg" alt="" width="435" height="267"></picture></div></div><p>&nbsp;</p><p>Let us consider a point Z such that normal distance OZ = 10 cm = 0.1 m</p><p>Let&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;and&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;be the electric potential and electric field respectively at Z.</p><p>From the&nbsp;<span title="Click to copy mathml"><math><mo>?</mo></math></span>&nbsp;ABZ, we can write,</p><p>BZ = AZ =&nbsp;<span title="Click to copy mathml"><math><msqrt><mrow><msup><mrow><mrow><mn>0.15</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mn>0.1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></msqrt></math></span>&nbsp;= 0.18 m</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi></mi></mrow></mrow></msub><mo>=</mo></math></span>&nbsp;Potential due to charge at A + Potential due to charge at B</p><p><span title="Click to copy mathml"><math><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></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><mi>A</mi><mi>Z</mi></mrow></mrow></mfrac></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></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><mi>B</mi><mi>Z</mi></mrow></mrow></mfrac></math></span>&nbsp;=&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><mn>0.18</mn></mrow></mrow></mfrac></math></span>&nbsp;(&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>+</mo><mi></mi><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>)</mo></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></math></span>&nbsp;= permittivity of free space = 8.854&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup></math></span>&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><msup><mrow><mrow><mi>N</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math></span>&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi></mi></mrow></mrow></msub><mo>=</mo></math></span>&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><mo>×</mo><mn>8.854</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup><mo>×</mo><mn>0.18</mn></mrow></mrow></mfrac></math></span>&nbsp;(1.5&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>+</mo><mi></mi><mn>2.5</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>)</mo></math></span>&nbsp;= 2.0&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;V</p><p>Electric field due to charge&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mi></mi><mi>a</mi><mi>t</mi><mi></mi><mi>Z</mi><mo>,</mo><mi></mi><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></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><mo>(</mo><msup><mrow><mrow><mi>A</mi><mi>Z</mi><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1.5</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4</mn><mo>×</mo><mi>π</mi><mo>×</mo><mn>8.854</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup><mo>×</mo><mo>(</mo><msup><mrow><mrow><mn>0.18</mn><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;= 0.416&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup></math></span>&nbsp;V/m</p><p>Electric field due to charge&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mi></mi><mi>a</mi><mi>t</mi><mi></mi><mi>Z</mi><mo>,</mo><mi></mi><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>q</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></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><mo>(</mo><msup><mrow><mrow><mi>B</mi><mi>Z</mi><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2.5</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup></mrow></mrow><mrow><mrow><mn>4</mn><mo>×</mo><mi>π</mi><mo>×</mo><mn>8.854</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>12</mn></mrow></mrow></msup><mo>×</mo><mo>(</mo><msup><mrow><mrow><mn>0.18</mn><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></mrow></mfrac></math></span>&nbsp;= 0.693&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup></math></span>&nbsp;V/m</p><p>The resultant intensity at Z</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><msqrt><mrow><msup><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><mn>2</mn><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>A</mi></mrow></mrow></msub><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mi>B</mi></mrow></mrow></msub><mi>c</mi><mi>o</mi><mi>s</mi><mn>2</mn><mi>θ</mi></mrow></msqrt></math></span>&nbsp;where 2&nbsp;<span title="Click to copy mathml"><math><mi>θ</mi></math></span>&nbsp;is the&nbsp;<span title="Click to copy mathml"><math><mi>∠</mi><mi>A</mi><mi>Z</mi><mi>B</mi></math></span></p><p>From the figure, we get&nbsp;<span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">cos</mi></mrow><mo>?</mo><mrow><mi>θ</mi></mrow></mrow></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>Z</mi><mi>O</mi></mrow></mrow><mrow><mrow><mi>Z</mi><mi>B</mi></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>0.1</mn></mrow></mrow><mrow><mrow><mn>0.18</mn></mrow></mrow></mfrac></math></span></p><p>Hence&nbsp;<span title="Click to copy mathml"><math><mi>θ</mi></math></span>&nbsp;= 56.25&nbsp;<span title="Click to copy mathml"><math><mo>°</mo></math></span></p><p>Therefore,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>=</mo><msqrt><mrow><mo>(</mo><msup><mrow><mrow><mn>0.416</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mo>(</mo><mn>0.693</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup><mo>)</mo></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><mn>2</mn><msub><mrow><mrow><mfenced separators="|"><mrow><mrow><mn>0.416</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup></mrow></mrow></mfenced><mo>×</mo><mo>(</mo><mn>0.693</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></msup><mo>)</mo></mrow></mrow><mrow><mrow></mrow></mrow></msub><mi>c</mi><mi>o</mi><mi>s</mi><mn>2</mn><mo>×</mo><mn>56.25</mn><mi mathvariant="normal"></mi><mo>°</mo></mrow></msqrt></math></span>&nbsp;= 6.57&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math></span>&nbsp;V/m</p></div></div>

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