Two point charges Q1=2μC and Q2=1μC are placed as shown. The coordinates of the point P are (2cm,1cm). The electric intensity vector at P subtends an angle θ with the positive X axis. The value of θ is given by:
A
tanθ=1
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B
tanθ=2
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C
tanθ=3
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D
tanθ=4
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Solution
The correct option is Atanθ=2
Given that:
Q1=2μC ans Q2=1μC
We know that:
E∝Qd2
⇒E1E2=2μ(2cm)2×(1cm)21μ=12
⇒E2=2E1
As →E1 is directed along +ve x-axis and →E2 along +ve y-axis, angle made by their resultant →E with +ve x-axis is given by tanθ=E2E1=2