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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 16 of 37
Marks: +1, -0
Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E⃗2−E⃗1)⋅n^(\vec{E}_2 - \vec{E}_1) \cdot \hat{n} = σε0\frac{\sigma}{\varepsilon_0} where n^\hat{n} is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of n^\hat{n} is from side 1 to side 2). Hence show that just outside a conductor, the electric field is σ n^ε0\frac{\hat{n}}{\varepsilon_0}
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Solution:  
Normal component of electric field intensity due to a thin infinite plane sheet of charge, on left side (side 1) E⃗1\vec{E}_1 = - σ2ε0n^\frac{\sigma}{2\varepsilon_0} \hat{n}
and on right side (side 2), E⃗2\vec{E}_2 = σ2ε0n^\frac{\sigma}{2\varepsilon_0} \hat{n}
Discontinuity in the normal component from one side to the other is
E⃗2−E⃗1\vec{E}_2 - \vec{E}_1 = σ2ε0n^+σ2ε0n^\frac{\sigma}{2\varepsilon_0} \hat{n} + \frac{\sigma}{2\varepsilon_0} \hat{n} = σε0n^\frac{\sigma}{\varepsilon_0} \hat{n} or (E⃗2−E⃗1)(\vec{E}_2 - \vec{E}_1) . n^\hat{n} = σε0n^⋅n^\frac{\sigma}{\varepsilon_0} \hat{n} \cdot \hat{n} = σε0\frac{\sigma}{\varepsilon_0}
Inside a closed conductor, E⃗1\vec{E}_1 = 0
∴ E = E⃗2\vec{E}_2 = σε0n^\frac{\sigma}{\varepsilon_0} \hat{n}
(b) To show that the tangential component of electrostatic field is continuous from one side of a charged surface to another, we use the fact that work done by electrostatic field on a closed loop is zero.
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