(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$({\mathit{E}}_2$ - ${\mathit{E}}_1).\widehat{\mathit{n}}=\frac{\mathit{\sigma }}{{\mathit{\epsilon }}_0}$

Where $\widehat{\mathit{n}}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of $\widehat{\mathit{n}}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is $\mathit{\sigma }\widehat{\mathit{n}}$ / ${\mathit{\epsilon }}_0$ .

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

<p><strong> 2.16 </strong>Electric field on one side of a charged body is&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;and electric field on the other side of the same body is&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><mi></mi></math></span>&nbsp;If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,</p><p><span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;………….(i)</p><p>Where,</p><p><span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;= unit vector normal to the surface at a point</p><p><span title="Click to copy mathml"><math><mi>σ</mi></math></span>&nbsp;= surface charge density at that point</p><p>Electric field due to the other surface of the charged body is given by</p><p><span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;………….(ii)</p><p><span title="Click to copy mathml"><math><mi mathvariant="normal">E</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">f</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">y</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">t</mi></math></span>&nbsp;due to the two surfaces,</p><p><span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow><mo>?</mo></mover><mo>-</mo><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow><mo>?</mo></mover><mo>=</mo><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover><mo>+</mo><mi></mi><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><mn>2</mn><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover><mo>=</mo><mi></mi><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;……(iii)</p><p><span title="Click to copy mathml"><math><mi mathvariant="normal">S</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">a</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">r</mi><mo>,</mo><mi mathvariant="normal"></mi><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi mathvariant="normal">E</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow><mo>?</mo></mover><mo>=</mo><mn>0</mn><mo>,</mo></math></span></p><p><span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><mi>E</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><msub><mrow><mrow><mi>E</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover></math></span></p><p>Therefore, the electric field just outside the conductor is&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>σ</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub></mrow></mrow></mfrac><mover accent="true"><mrow><mrow><mi>n</mi></mrow></mrow><mo>?</mo></mover></math></span></p><p>When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.</p>

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