A) \[\frac{{{q}^{2}}Sd}{8{{\pi }^{2}}{{\varepsilon }_{0}}{{\ell }^{5}}}\]
B) \[\frac{{{q}^{2}}Sd}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{\ell }^{5}}}\]
C) \[\frac{{{q}^{2}}Sd}{6{{\pi }^{2}}{{\varepsilon }_{0}}{{\ell }^{5}}}\]
D) \[\frac{2{{q}^{2}}Sd}{3{{\pi }^{2}}{{\varepsilon }_{0}}{{\ell }^{5}}}\]
Correct Answer: A
Solution :
[a] Since the sheet is metallic, the charges must Be redistributed over its surface so that the field in the bulk of the sheet is zero. In the first approximation, we can assume that this distribution is uniform and has density \[-\,\alpha \]and \[\sigma \]over the upper and the lower surface respectively of the sheet. According to the superposition principle, we obtain the condition for the absence of the field in the bulk of the sheet: \[\frac{q}{4\pi {{\varepsilon }_{0}}{{\ell }^{2}}}-\frac{\sigma }{{{\varepsilon }_{0}}}=0\] Let us now take into consideration the non-uniformity of the field produced by the point charge since it is the single cause of the force F of integration. The upper surface of the sheet must be attracted with a force \[\frac{\sigma sq}{4\pi {{\varepsilon }_{0}}{{\ell }^{2}}},\] while the lower surface must be repelled with a force \[F=\frac{\sigma Sq}{4\pi {{\varepsilon }_{0}}{{\ell }^{2}}}\left[ 1-\frac{1}{{{\left( 1+d/\ell \right)}^{2}}} \right]\approx \frac{{{q}^{2}}Sd}{8{{\pi }^{2}}{{\varepsilon }_{0}}{{\ell }^{5}}}\]
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