Question:
Four charges, each charge \(q\) coulomb are placed at points \((-1,0,0)\), \((1,0,0)\), \((0,-1,0)\), and \((0,1,0)\) in the \(xy\)-plane. The distances along the axes are measured in meters. The magnitude of the electric field at the point \(z=1\,\text{m}\) on the \(z\)-axis is
Four charges, each charge \(q\) coulomb are placed at points \((-1,0,0)\), \((1,0,0)\), \((0,-1,0)\), and \((0,1,0)\) in the \(xy\)-plane. The distances along the axes are measured in meters. The magnitude of the electric field at the point \(z=1\,\text{m}\) on the \(z\)-axis is
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For symmetric charge arrangements, resolve electric fields into components. The horizontal components often cancel, while axial components add.
Updated On: Jun 26, 2026
- \(\frac{1}{2\sqrt{2}}\frac{q}{\pi\varepsilon_0}\,\text{C m}^{-2}\)
- \(\frac{1}{4}\frac{q}{\pi\varepsilon_0}\,\text{C m}^{-2}\)
- \(\frac{q}{\pi\varepsilon_0}\,\text{C m}^{-2}\)
- \(\frac{q}{2\pi\varepsilon_0}\,\text{C m}^{-2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the point where electric field is required.
The point on the \(z\)-axis is \[ P(0,0,1) \] The four charges are at \[ (-1,0,0),\quad (1,0,0),\quad (0,-1,0),\quad (0,1,0) \] Each charge is at the same distance from \(P\).
Step 2: Find the distance from each charge to \(P\).
For the charge at \((1,0,0)\), the distance from \(P(0,0,1)\) is \[ r=\sqrt{(0-1)^2+(0-0)^2+(1-0)^2} \] \[ r=\sqrt{1+1} \] \[ r=\sqrt{2} \] So, each charge is at distance \[ \sqrt{2}\,\text{m} \] from \(P\).
Step 3: Find electric field due to one charge.
Electric field due to one charge is \[ E_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \] Since \[ r^2=2, \] we get \[ E_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{2} \] \[ E_1=\frac{q}{8\pi\varepsilon_0} \]
Step 4: Find the \(z\)-component of electric field due to one charge.
The line joining each charge to \(P\) makes an angle \(\theta\) with the \(z\)-axis.
Here, \[ \cos\theta=\frac{\text{vertical distance}}{\text{distance}} \] \[ \cos\theta=\frac{1}{\sqrt{2}} \] Therefore, the \(z\)-component due to one charge is \[ E_{1z}=E_1\cos\theta \] \[ E_{1z}=\frac{q}{8\pi\varepsilon_0}\cdot \frac{1}{\sqrt{2}} \] \[ E_{1z}=\frac{q}{8\sqrt{2}\pi\varepsilon_0} \]
Step 5: Add the components due to all four charges.
Due to symmetry, the \(x\)- and \(y\)-components cancel each other.
Only the \(z\)-components add.
Thus, \[ E=4E_{1z} \] \[ E=4\cdot \frac{q}{8\sqrt{2}\pi\varepsilon_0} \] \[ E=\frac{q}{2\sqrt{2}\pi\varepsilon_0} \] \[ E=\frac{1}{2\sqrt{2}}\frac{q}{\pi\varepsilon_0} \]
Step 6: Final conclusion.
Hence, the magnitude of the electric field is \[ \boxed{\frac{1}{2\sqrt{2}}\frac{q}{\pi\varepsilon_0}} \]
The point on the \(z\)-axis is \[ P(0,0,1) \] The four charges are at \[ (-1,0,0),\quad (1,0,0),\quad (0,-1,0),\quad (0,1,0) \] Each charge is at the same distance from \(P\).
Step 2: Find the distance from each charge to \(P\).
For the charge at \((1,0,0)\), the distance from \(P(0,0,1)\) is \[ r=\sqrt{(0-1)^2+(0-0)^2+(1-0)^2} \] \[ r=\sqrt{1+1} \] \[ r=\sqrt{2} \] So, each charge is at distance \[ \sqrt{2}\,\text{m} \] from \(P\).
Step 3: Find electric field due to one charge.
Electric field due to one charge is \[ E_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} \] Since \[ r^2=2, \] we get \[ E_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{2} \] \[ E_1=\frac{q}{8\pi\varepsilon_0} \]
Step 4: Find the \(z\)-component of electric field due to one charge.
The line joining each charge to \(P\) makes an angle \(\theta\) with the \(z\)-axis.
Here, \[ \cos\theta=\frac{\text{vertical distance}}{\text{distance}} \] \[ \cos\theta=\frac{1}{\sqrt{2}} \] Therefore, the \(z\)-component due to one charge is \[ E_{1z}=E_1\cos\theta \] \[ E_{1z}=\frac{q}{8\pi\varepsilon_0}\cdot \frac{1}{\sqrt{2}} \] \[ E_{1z}=\frac{q}{8\sqrt{2}\pi\varepsilon_0} \]
Step 5: Add the components due to all four charges.
Due to symmetry, the \(x\)- and \(y\)-components cancel each other.
Only the \(z\)-components add.
Thus, \[ E=4E_{1z} \] \[ E=4\cdot \frac{q}{8\sqrt{2}\pi\varepsilon_0} \] \[ E=\frac{q}{2\sqrt{2}\pi\varepsilon_0} \] \[ E=\frac{1}{2\sqrt{2}}\frac{q}{\pi\varepsilon_0} \]
Step 6: Final conclusion.
Hence, the magnitude of the electric field is \[ \boxed{\frac{1}{2\sqrt{2}}\frac{q}{\pi\varepsilon_0}} \]
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