Question

An infinitely large non-conducting thin sheet in the \(xy\) plane (\(z=0\)) carries a uniform surface charge density \(\sigma = 17.70\times10^{-12}\) C.m\(^{-2}\). The electric field in the region \(z < 0\) is \(\vec{E}_2 = \hat{x}+2\hat{y}+3\hat{z}\). Then, the electric field \(\vec{E}_1\) in the region \(z > 0\) will be (\(\epsilon_0 = 8.85\times10^{-12}\) C\(^2\).N\(^{-1}\).m\(^{-2}\))

• A. \(\vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z}\)
• B. \(\vec{E}_1 = \hat{x}+2\hat{y}+4\hat{z}\)
• C. \(\vec{E}_1 = \hat{x}+2\hat{y}+3\hat{z}\)
• D. \(\vec{E}_1 = \hat{x}+4\hat{y}+\hat{z}\)

Hint:

Tangential components of \(\vec{E}\) are continuous across a charged sheet; only the normal component jumps, by \(\sigma/\epsilon_0\).

Answer 1

The Correct Option is A

Step 1: Build a thin pillbox across the sheet.
Take a small pillbox straddling the sheet at $z=0$, with its two flat faces (area $A$) parallel to the sheet, one at $z=+\delta$ in region 1 and one at $z=-\delta$ in region 2, with $\delta\to 0$. Gauss's law says the total outward flux through the pillbox equals the enclosed charge divided by $\epsilon_0$.

Step 2: Apply Gauss's law to the pillbox.
The side walls shrink to zero area as $\delta\to0$, so only the two flat faces contribute flux. The outward normal on the top face is $+\hat{z}$ and on the bottom face is $-\hat{z}$:
\[ E_{1z}A - E_{2z}A = \frac{\sigma A}{\epsilon_0} \]
Cancelling $A$ gives $E_{1z}-E_{2z} = \sigma/\epsilon_0$, the jump in the direction normal to the sheet.

Step 3: Show the tangential parts cannot jump.
Now take a thin rectangular loop straddling the sheet, with its long sides (length $L$) lying along $\hat{x}$, one just above and one just below $z=0$, joined by short sides of length $2\delta\to0$. Since the fields here are static, $\oint \vec{E}\cdot d\vec{l} = 0$ around any closed loop. The short sides contribute nothing as $\delta\to0$, so the two long sides must cancel: $E_{1x}L - E_{2x}L = 0$, giving $E_{1x}=E_{2x}$. The same loop with sides along $\hat{y}$ gives $E_{1y}=E_{2y}$.

Step 4: Plug in the numbers.
From $\vec{E}_2 = \hat{x}+2\hat{y}+3\hat{z}$: $E_{2x}=1$, $E_{2y}=2$, $E_{2z}=3$. By Step 3, $E_{1x}=1$ and $E_{1y}=2$. By Step 2, $E_{1z} = E_{2z}+\sigma/\epsilon_0$. Compute the ratio:
\[ \frac{\sigma}{\epsilon_0} = \frac{17.70\times10^{-12}}{8.85\times10^{-12}} = 2 \]
So $E_{1z} = 3+2 = 5$.

Step 5: Assemble the answer.
\[ \vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z} \]
This matches option (A). The pillbox fixes the perpendicular jump and the flat loop confirms the parallel parts cannot change, so both halves of the boundary condition come out from first principles rather than a memorized rule.

Final Answer:
Only the field component perpendicular to the charged sheet jumps, by $\sigma/\epsilon_0$. \[ \boxed{\vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z}} \]