Question:
Three charges 1 C, 2 C and 3 C are placed at points (1, 0, 0), (0, 2, 0) and (0, 0, 3) respectively. The distances are measured in metre along the three axes. The magnitude of the electric field at the origin is:
Three charges 1 C, 2 C and 3 C are placed at points (1, 0, 0), (0, 2, 0) and (0, 0, 3) respectively. The distances are measured in metre along the three axes. The magnitude of the electric field at the origin is:
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Electric field components from perpendicular axes can be added using Pythagoras theorem.
Updated On: Jun 20, 2026
- \(\frac{7}{24\pi \varepsilon_0}\)
- \(\frac{11}{24\pi \varepsilon_0}\)
- \(\frac{49}{144\pi \varepsilon_0}\)
- \(\frac{25}{144\pi \varepsilon_0}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand electric field due to point charge.
Electric field due to a point charge: \[ E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2} \] Direction is away from positive charge and along the line joining charge to observation point.
Step 2: Electric field due to charge at (1,0,0).
Charge \(q_1 = 1C\), distance from origin = 1 m. \[ E_1 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{1}{1^2} = \frac{1}{4\pi \varepsilon_0} \] Direction: along negative x-axis. So, \[ \vec{E_1} = -\frac{1}{4\pi \varepsilon_0}\hat{i} \]
Step 3: Electric field due to charge at (0,2,0).
Charge \(q_2 = 2C\), distance = 2 m. \[ E_2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2}{4} = \frac{1}{8\pi \varepsilon_0} \] Direction: negative y-axis. \[ \vec{E_2} = -\frac{1}{8\pi \varepsilon_0}\hat{j} \]
Step 4: Electric field due to charge at (0,0,3).
Charge \(q_3 = 3C\), distance = 3 m. \[ E_3 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{3}{9} = \frac{1}{12\pi \varepsilon_0} \] Direction: negative z-axis. \[ \vec{E_3} = -\frac{1}{12\pi \varepsilon_0}\hat{k} \]
Step 5: Resultant electric field magnitude.
\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \] \[ = \frac{1}{4\pi \varepsilon_0} \sqrt{1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2} \]
Step 6: Simplify expression.
\[ = \frac{1}{4\pi \varepsilon_0} \sqrt{1 + \frac{1}{4} + \frac{1}{9}} \] LCM = 36: \[ = \frac{1}{4\pi \varepsilon_0} \sqrt{\frac{36 + 9 + 4}{36}} \] \[ = \frac{1}{4\pi \varepsilon_0} \sqrt{\frac{49}{36}} \] \[ = \frac{1}{4\pi \varepsilon_0} \cdot \frac{7}{6} \]
Step 7: Final answer.
\[ \boxed{\frac{7}{24\pi \varepsilon_0}} \]
Electric field due to a point charge: \[ E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2} \] Direction is away from positive charge and along the line joining charge to observation point.
Step 2: Electric field due to charge at (1,0,0).
Charge \(q_1 = 1C\), distance from origin = 1 m. \[ E_1 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{1}{1^2} = \frac{1}{4\pi \varepsilon_0} \] Direction: along negative x-axis. So, \[ \vec{E_1} = -\frac{1}{4\pi \varepsilon_0}\hat{i} \]
Step 3: Electric field due to charge at (0,2,0).
Charge \(q_2 = 2C\), distance = 2 m. \[ E_2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2}{4} = \frac{1}{8\pi \varepsilon_0} \] Direction: negative y-axis. \[ \vec{E_2} = -\frac{1}{8\pi \varepsilon_0}\hat{j} \]
Step 4: Electric field due to charge at (0,0,3).
Charge \(q_3 = 3C\), distance = 3 m. \[ E_3 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{3}{9} = \frac{1}{12\pi \varepsilon_0} \] Direction: negative z-axis. \[ \vec{E_3} = -\frac{1}{12\pi \varepsilon_0}\hat{k} \]
Step 5: Resultant electric field magnitude.
\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \] \[ = \frac{1}{4\pi \varepsilon_0} \sqrt{1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2} \]
Step 6: Simplify expression.
\[ = \frac{1}{4\pi \varepsilon_0} \sqrt{1 + \frac{1}{4} + \frac{1}{9}} \] LCM = 36: \[ = \frac{1}{4\pi \varepsilon_0} \sqrt{\frac{36 + 9 + 4}{36}} \] \[ = \frac{1}{4\pi \varepsilon_0} \sqrt{\frac{49}{36}} \] \[ = \frac{1}{4\pi \varepsilon_0} \cdot \frac{7}{6} \]
Step 7: Final answer.
\[ \boxed{\frac{7}{24\pi \varepsilon_0}} \]
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