Question

the electric field at a distance of 20cm from 2uc charge is....
explain???​

Answer 1

Electrostatic Field Intensity :

$E = \dfrac{kq}{ {r}^{2} }$

• 'k' is Coulomb's Constant and equal to 9 × 10⁹.
• "q" is charge and "r" is distance.

$\implies E = \dfrac{(9 \times {10}^{9}) \times (2 \times {10}^{ - 6}) }{ {(0.2)}^{2} }$

$\implies E = \dfrac{18 \times {10}^{3} }{ 0.04}$

$\implies E = 4.5 \times {10}^{5} \: N/C$

So, field intensity is 4.5 × 10⁵ N/C