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Difficulty: MediumElectrostatics and Electric Charges

Two point charges produce mutually perpendicular electric fields at a point PP in a vacuum. If the magnitudes of the electric field intensities at PP due to the charges individually are 30 N C−130\text{ N C}^{-1} and 40 N C−140\text{ N C}^{-1}, what is the magnitude of the net electric field intensity at point PP?

  1. 50 N C−150\text{ N C}^{-1}Answer
  2. B
    70 N C−170\text{ N C}^{-1}
  3. C
    10 N C−110\text{ N C}^{-1}
  4. D
    35 N C−135\text{ N C}^{-1}

Answer

The magnitude of the net electric field intensity at point PP is 50 N C−150\text{ N C}^{-1}.
Because electric field intensity is a vector quantity, two mutually perpendicular electric fields E1=30 N C−1E_1 = 30\text{ N C}^{-1} and E2=40 N C−1E_2 = 40\text{ N C}^{-1} combine vectorially according to Enet=E12+E22=302+402=50 N C−1E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{30^2 + 40^2} = 50\text{ N C}^{-1}.

Step-by-Step Solution

1
Identify the vector nature and orientation of the given electric fields.
The two component fields E1=30 N C−1E_1 = 30\text{ N C}^{-1} and E2=40 N C−1E_2 = 40\text{ N C}^{-1} are perpendicular to each other (θ=90∘\theta = 90^\circ).
Electric field intensity is a vector quantity, so perpendicular vectors must be combined using vector addition.
2
Apply the Pythagorean theorem to calculate the resultant vector magnitude.
Enet=E12+E22=302+402=900+1600=2500=50 N C−1E_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\text{ N C}^{-1}.
When two vectors meet at a right angle, the magnitude of their resultant is the hypotenuse of the right-angled triangle formed by the vector components.

Key Concept

Vector Addition of Electric Fields (Superposition Principle)
Estimated Time:1m 0s
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Topics

Electrostatics and Electric Charges

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