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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:
476494_addf54ad897a49fabfd1d38e12149b5c.png


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 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 $$\tan { \theta  } =\dfrac { { E }_{ 2 } }{ { E }_{ 1 } } =2$$

512895_476494_ans_eba4dcca768647d886f18df8008e6668.png

Physics
NCERT
Standard XII

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