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

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

$E = 9 \times 10^{9} \times \frac{Q}{r^{2}} \Rightarrow 500 = 9 \times 10^{9} \times \frac{Q}{(3)^{2}} \Rightarrow Q = 0.5 μ C$

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

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 $[\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} / C^{2}]$ :

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The electric field due to a charge at a distance of 3 m from it is 500 N/coulomb. The magnitude of the charge is [14πϵ0=9×109 N−m2coulomb2]

The correct option is D 0.5 micro-coulomb E=9×109×Qr2⇒500=9×109×Q(3)2⇒Q=0.5μC

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

The correct option is D
0.5 micro-coulomb

$E = 9 \times 10^{9} \times \frac{Q}{r^{2}} \Rightarrow 500 = 9 \times 10^{9} \times \frac{Q}{(3)^{2}} \Rightarrow Q = 0.5 μ C$