A point charge of 100-C is placed at 3 i - 4 j m- Find the electric field intensity due to this charge at a point located at 9 i - 12 j m -

The correct option is B 9000 Vm^ 1 #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; ...

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Byju's Answer

Standard XII

Physics

Electric Field

A point charg...

Question

A point charge of $100 μ$ C is placed at $3 ˆ i + 4 ˆ j$ m. Find the electric field intensity due to this charge at a point located at $9 ˆ i + 12 ˆ j m$ :

800 V m^{- 1}

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9000 V m^{- 1}

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2250 V m^{- 1}

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4500 V m^{- 1}

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Solution

The correct option is B $9000 V m^{- 1}$

$\to r = (9 - 3) ˆ i + (12 - 4) ˆ j = 6 ˆ i + 8 ˆ j$

or $r = \sqrt{6^{2} + 8^{2}} = 10 m$

$E = \frac{K q}{r^{2}}$

Electric field intensity due to point charge $E = \frac{9 \times 10^{9} \times 100 \times 10^{- 6}}{10^{2}} = 9000 V m^{- 1}$

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A point charge of 100μC is placed at 3ˆi+4ˆj m. Find the electric field intensity due to this charge at a point located at 9ˆi+12ˆj m :

The correct option is B 9000Vm−1 →r=(9−3)ˆi+(12−4)ˆj=6ˆi+8ˆj or r=√62+82=10m E=Kqr2 Electric field intensity due to point charge E=9×109×100×10−6102=9000Vm−1

A point charge of $100 μ$ C is placed at $3 ˆ i + 4 ˆ j$ m. Find the electric field intensity due to this charge at a point located at $9 ˆ i + 12 ˆ j m$ :

The correct option is B $9000 V m^{- 1}$

$\to r = (9 - 3) ˆ i + (12 - 4) ˆ j = 6 ˆ i + 8 ˆ j$

or $r = \sqrt{6^{2} + 8^{2}} = 10 m$

$E = \frac{K q}{r^{2}}$

Electric field intensity due to point charge $E = \frac{9 \times 10^{9} \times 100 \times 10^{- 6}}{10^{2}} = 9000 V m^{- 1}$