Question

The value of ${\epsilon }_0$ will be

• A. $8.85\times {10}^{-12}{C}^2/N-m^2$
• B. $8.85\times {10}^{-12}Fermi$
• C. $9\times {10}^{-13}{C}^2/N-m^2$
• D. $9\times {10}^{-13}Fermi$

Answer 1

Answer: (A)

$\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2/C$
$\therefore$ ${\epsilon }_0=\frac{1}{9\times {10}^9\times 4\pi }{C}^2/N-m^2$
$=8.85\times {10}^{-12}{C}^2/N-m^2$