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\(^{-2}\))
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\(^{-2}\))
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Tangential components of \(\vec{E}\) are continuous across a charged sheet; only the normal component jumps, by \(\sigma/\epsilon_0\).
Updated On: Jul 28, 2026
- \(\vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z}\)
- \(\vec{E}_1 = \hat{x}+2\hat{y}+4\hat{z}\)
- \(\vec{E}_1 = \hat{x}+2\hat{y}+3\hat{z}\)
- \(\vec{E}_1 = \hat{x}+4\hat{y}+\hat{z}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
A surface charge sitting on a boundary produces a fixed jump in the electric field across that boundary. The component of \(\vec{E}\) parallel to the sheet (tangential) stays the same on both sides, while the component perpendicular to the sheet (normal) jumps by \(\sigma/\epsilon_0\) as we cross the sheet.
Step 2: Key Formula or Approach:
For a sheet in the \(xy\)-plane with the normal direction taken as \(\hat{z}\), pointing from the \(z<0\) side into the \(z>0\) side:
1. Tangential components are continuous: \(E_{1x} = E_{2x}\), \(E_{1y}=E_{2y}\).
2. Normal component jumps: \(E_{1z} - E_{2z} = \sigma/\epsilon_0\).
Step 3: Detailed Explanation:
Read off the given field for \(z<0\): \(\vec{E}_2 = \hat{x}+2\hat{y}+3\hat{z}\), so \(E_{2x}=1\), \(E_{2y}=2\), \(E_{2z}=3\).
Since the sheet carries no surface current, the \(x\) and \(y\) components of \(\vec{E}\) do not change across it:
\[ E_{1x} = 1, \qquad E_{1y} = 2 \]
For the normal (\(z\)) component, compute \(\sigma/\epsilon_0\):
\[ \frac{\sigma}{\epsilon_0} = \frac{17.70\times10^{-12}}{8.85\times10^{-12}} = 2 \]
So the \(z\)-component increases by 2 on crossing from \(z<0\) to \(z>0\):
\[ E_{1z} = E_{2z} + \frac{\sigma}{\epsilon_0} = 3 + 2 = 5 \]
Putting the three components together:
\[ \vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z} \]
Step 4: Why the other options are wrong.
Option (B) uses \(E_{1z}=4\), as if only half the jump were added, the kind of slip that comes from mixing up this boundary-condition jump with the separate formula \(\sigma/2\epsilon_0\) used for the field of an isolated sheet in open space. Option (C) keeps \(\vec{E}_1\) equal to \(\vec{E}_2\), ignoring the charge on the sheet altogether. Option (D) changes the tangential components instead of the normal one, which contradicts the rule that only the normal component of \(\vec{E}\) can jump across a surface charge.
Final Answer:
The tangential parts of the field stay fixed and only the normal part jumps, by \(\sigma/\epsilon_0=2\). \[ \boxed{\vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z}} \]
A surface charge sitting on a boundary produces a fixed jump in the electric field across that boundary. The component of \(\vec{E}\) parallel to the sheet (tangential) stays the same on both sides, while the component perpendicular to the sheet (normal) jumps by \(\sigma/\epsilon_0\) as we cross the sheet.
Step 2: Key Formula or Approach:
For a sheet in the \(xy\)-plane with the normal direction taken as \(\hat{z}\), pointing from the \(z<0\) side into the \(z>0\) side:
1. Tangential components are continuous: \(E_{1x} = E_{2x}\), \(E_{1y}=E_{2y}\).
2. Normal component jumps: \(E_{1z} - E_{2z} = \sigma/\epsilon_0\).
Step 3: Detailed Explanation:
Read off the given field for \(z<0\): \(\vec{E}_2 = \hat{x}+2\hat{y}+3\hat{z}\), so \(E_{2x}=1\), \(E_{2y}=2\), \(E_{2z}=3\).
Since the sheet carries no surface current, the \(x\) and \(y\) components of \(\vec{E}\) do not change across it:
\[ E_{1x} = 1, \qquad E_{1y} = 2 \]
For the normal (\(z\)) component, compute \(\sigma/\epsilon_0\):
\[ \frac{\sigma}{\epsilon_0} = \frac{17.70\times10^{-12}}{8.85\times10^{-12}} = 2 \]
So the \(z\)-component increases by 2 on crossing from \(z<0\) to \(z>0\):
\[ E_{1z} = E_{2z} + \frac{\sigma}{\epsilon_0} = 3 + 2 = 5 \]
Putting the three components together:
\[ \vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z} \]
Step 4: Why the other options are wrong.
Option (B) uses \(E_{1z}=4\), as if only half the jump were added, the kind of slip that comes from mixing up this boundary-condition jump with the separate formula \(\sigma/2\epsilon_0\) used for the field of an isolated sheet in open space. Option (C) keeps \(\vec{E}_1\) equal to \(\vec{E}_2\), ignoring the charge on the sheet altogether. Option (D) changes the tangential components instead of the normal one, which contradicts the rule that only the normal component of \(\vec{E}\) can jump across a surface charge.
Final Answer:
The tangential parts of the field stay fixed and only the normal part jumps, by \(\sigma/\epsilon_0=2\). \[ \boxed{\vec{E}_1 = \hat{x}+2\hat{y}+5\hat{z}} \]
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