What is the magnitude of the electric field at a point $2.0 m$ from a point charge $q = 4.0 n C$ ?

9 N / C

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90 N / C

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0.9 N / C

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900 N / C

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Solution

The correct option is A $9 N / C$
$E = \frac{k q}{r^{2}}$

$⟹ E = \frac{9 \times 10^{9} \times 4 \times 10^{- 9}}{2^{2}}$

$⟹ E = 9 N / C$

Answer-(A)

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What is the magnitude of the electric field at a point 2.0m from a point charge q=4.0nC?

The correct option is A 9N/CE=kqr2⟹E=9×109×4×10−922⟹E=9N/CAnswer-(A)

What is the magnitude of the electric field at a point $2.0 m$ from a point charge $q = 4.0 n C$ ?

The correct option is A $9 N / C$
$E = \frac{k q}{r^{2}}$

$⟹ E = \frac{9 \times 10^{9} \times 4 \times 10^{- 9}}{2^{2}}$

$⟹ E = 9 N / C$

Answer-(A)