Question

Calculate the value of the electric field when the field point is located at 3cm from the point charge.The charge of the point charge is 4nC. (A) 4000N/C (B) 2000N/C (C) 3300N/C (D) 1000N/C

Answer 1

Answer:

40000 N/C

Explanation:

$\overrightarrow{E}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q}{r^{2}}$

Here, r = 3 cm = $3\times 10^{-2}m.$ And $q=4\times 10^{-9}C$.

We know that in SI units, the value of

$\dfrac{1}{4\pi \varepsilon _{0}}=9\times 10^{9}Nm^{2}/C^{2}$

Therefore,

$\overrightarrow{E}=9\times 10^{9}Nm^{2}\right| C^{-2}\times \dfrac{4\times 10^{-9}C}{\left( 3\times 10^{-2}m\right)^{2}}\end{aligned}$

$=9\times 10^{9}\times \dfrac{4\times 10^{-9}}{9\times 10^{-4}}\dfrac{N}{C}$

$=4\times 10^{4}N/C$

=40000 N/C