Discussion
An infinitely large non-conducting thin sheet in the \(xy\) plane (\(z\)=0) carries a uniform surface charge density \(\sigma = 17.70 \times 10^{-12}\) C.m−2. The electric field in the region \(z \lt 0\) is \(\vec{E}_{2} = \hat{x} + 2\hat{y} + 3\hat{z}\). Then, the electric field \(\vec{E}_{1}\) in the region \(z \gt 0\) will be
(\(\epsilon_{0} = 8.85 \times 10^{-12}\) C2.N−1.m−2)
Answer: A
The field above the sheet is \(\vec{E}_{1} = \hat{x} + 2\hat{y} + 5\hat{z}\).
At a charged surface the tangential component of \(\vec{E}\) is continuous, and the normal component jumps by \(\sigma/\epsilon_{0}\):
\[ \begin{aligned} E_{1z} - E_{2z} &= \frac{\sigma}{\epsilon_{0}} = \frac{17.70 \times 10^{-12}}{8.85 \times 10^{-12}} = 2 \\ E_{1z} &= 3 + 2 = 5 \end{aligned} \]
The x and y components, 1 and 2, carry over unchanged. Option with \(4\hat{z}\) adds only \(\sigma/2\epsilon_{0}\), the field of the sheet alone on one side; \(3\hat{z}\) ignores the charge, and the last option changes the tangential components, which a surface charge cannot do.
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