Discussion
Three infinite plane sheets carrying uniform charge densities \(-\sigma, 2\sigma, 3\sigma\) are placed parallel to the x-z plane at \(y = a, 3a, 4a\), respectively. The electric field at the point \((0, 2a, 0)\) is
Answer: B
The field is \(-\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath}\).
An infinite sheet of charge density \(\sigma\) gives a field \(\frac{\sigma}{2\varepsilon_{0}}\) pointing away from it (towards it for negative charge). The point \(y = 2a\) lies above the sheet at \(y = a\) and below the sheets at \(3a\) and \(4a\):
- \(-\sigma\) at \(y = a\): field points towards the sheet, \(-\frac{\sigma}{2\varepsilon_{0}}\hat{\jmath}\).
- \(2\sigma\) at \(y = 3a\): points away, downwards, \(-\frac{2\sigma}{2\varepsilon_{0}}\hat{\jmath}\).
- \(3\sigma\) at \(y = 4a\): points away, downwards, \(-\frac{3\sigma}{2\varepsilon_{0}}\hat{\jmath}\).
\[ \begin{aligned} \vec{E} &= -\frac{(1 + 2 + 3)\sigma}{2\varepsilon_{0}}\hat{\jmath} \\ &= -\frac{3\sigma}{\varepsilon_{0}}\hat{\jmath} \end{aligned} \]
Getting the sign of one sheet wrong gives \(-\frac{2\sigma}{\varepsilon_{0}}\) or \(\frac{\sigma}{\varepsilon_{0}}\).
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