Electrostatics Question 405

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 $ [ \frac{1}{4\pi {\varepsilon _{0}}}=9\times 10^{9}\frac{N-m^{2}}{coulomb^{2}} ] $ [MP PMT 2000]

Options:

A) 2.5 micro-coulomb

B) 2.0 micro-coulomb

C) 1.0 micro-coulomb

D) 0.5 micro-coulomb

Show Answer

Answer:

Correct Answer: D

Solution:

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

therefore Q = 0.5 mC



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