Question

Two point charges $${ Q }_{ 1 }=2\mu C$$ and $${ Q }_{ 2 }=1\mu C$$ are placed as shown. The coordinates of the point $$P$$ are $$\left( 2cm,1cm \right)$$. The electric intensity vector at $$P$$ subtends an angle $$\theta$$ with the positive $$X$$ axis. The value of $$\theta$$ is given by:

A
tanθ=1
B
tanθ=2
C
tanθ=3
D
tanθ=4

Solution

The correct option is A $$\tan { \theta } =2$$Given that: $$Q_1=2\ \mu C$$ ans $$Q_2=1\ \mu C$$We know that: $$E\propto \dfrac { Q }{ { d }^{ 2 } }$$$$\Rightarrow \dfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\dfrac { 2\mu }{ { \left( 2cm \right) }^{ 2 } } \times \dfrac { { \left( 1cm \right) }^{ 2 } }{ 1\mu } =\dfrac { 1 }{ 2 }$$$$\Rightarrow { E }_{ 2 }=2{ E }_{ 1 }$$As $$\vec { { E }_{ 1 } }$$ is directed along +ve x-axis and $$\vec { { E }_{ 2 } }$$ along +ve y-axis, angle made by their resultant $$\vec { E }$$ with +ve x-axis is given by $$\tan { \theta } =\dfrac { { E }_{ 2 } }{ { E }_{ 1 } } =2$$PhysicsNCERTStandard XII

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