Question

$\huge{\mathfrak{quEstion:-}}$
Find Electric Field at a distance 1m from charge 1μC ​

Answer 1

Answer:

$\large\boxed{\sf{9000\;\:N{C}^{-1}}}$

Step-by-step explanation:

Given that,

• Magnitude of Charge, q = 1 μC = ${10}^{-6}\;C$
• Distance, r = 1 m

To find the electric field.

We know that, electric field is given by,

• $\large{\red{E=\dfrac{1}{4\pi {\epsilon}_{0}}\dfrac{q}{{r}^{2}}}}$

Where, $\dfrac{1}{4\pi {\epsilon}_{0}}=9\times{10}^{9}\:N{m}^{2}{C}^{-2}$

Substituting the values, we get,

$= > E = 9 \times {10}^{9} \times \dfrac{ {10}^{ - 6} }{ {1}^{2} } \\ \\ = > E = 9 \times {10}^{9 - 6} \\ \\ = > E = 9 \times {10}^{3} \\ \\ = > E = 9000$

Hence, the required magnitude of electric field is 9000 N/C