Question

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 $\left\lbrack \frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}\frac{N - m^{2}}{coulomb^{2}} \right\rbrack$

• A. 2.5 micro-coulomb
• B. 2.0 micro-coulomb
• C. 1.0 micro-coulomb
• D. 0.5 micro-coulomb

Answer 1

Answer: (D)

0.5 micro-coulomb

Answer 2

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