The electric field due to a charge at a distance of $3 m$ from it is $500 N / C$ . The magnitude of the charge is $[\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} / C^{2}]$ :

2.5

micro-coulomb

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2.0

micro-coulomb

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1.0

micro-coulomb

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0.5

micro-coulomb

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Solution

The correct option is D $0.5$ micro-coulomb
We know $E$ due to charge $q$ $= \frac{k q}{r^{2}}$
$E = \frac{k q}{r^{2}}$
$\Rightarrow 500 = \frac{(9 \times 10^{9}) \times q}{(3)^{2}}$

$\Rightarrow q = 5 \times 10^{- 7} C$
$\Rightarrow q = 0 \cdot 5 μ C$

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The electric field due to a charge at a distance of 3m from it is 500N/C. The magnitude of the charge is[14πε0=9×109Nm2/C2] :

The correct option is D 0.5 micro-coulombWe know E due to charge q =kqr2E=kqr2⇒500=(9×109)×q(3)2⇒q=5×10−7C⇒q=0⋅5μC

The electric field due to a charge at a distance of $3 m$ from it is $500 N / C$ . The magnitude of the charge is $[\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} / C^{2}]$ :

The correct option is D $0.5$ micro-coulomb
We know $E$ due to charge $q$ $= \frac{k q}{r^{2}}$
$E = \frac{k q}{r^{2}}$
$\Rightarrow 500 = \frac{(9 \times 10^{9}) \times q}{(3)^{2}}$

$\Rightarrow q = 5 \times 10^{- 7} C$
$\Rightarrow q = 0 \cdot 5 μ C$