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 [ 1 4 π ϵ 0 = 9 × 10 9 N − m 2 c o u l o m b 2 ] Options: A . &nbsp 2.5 micro-coulomb B . &nbsp 2.0 micro-coulomb C . &nbsp 1.0 micro-coulomb D . &nbsp 0.5 micro-coulomb

Answer: Option D : D E = 9 × 10 9 × Q r 2 ⇒ 500 = 9 × 10 9 × Q ( 3 ) 2 ⇒ Q = 0.5 μ C

The electric field due to a charge at a distance of 3 m from it is 500 N/coulo

Discussion Forum : Electric Charges Fields And Gauss Law

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 $[\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N - m^{2}}{c o u l o m b^{2}}]$

Options:

A . 2.5 micro-coulomb

B . 2.0 micro-coulomb

C . 1.0 micro-coulomb

D . 0.5 micro-coulomb

Answer: Option D
:
D

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

Was this answer helpful ?

Next Question

Discussion Forum : Electric Charges Fields And Gauss Law

Submit Your Solution hear: