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1. Electric Charges and Fields
easy
An infinite line charge produce a field of $7.182 \times {10^8}\,N/C$ at a distance of $2\, cm$. The linear charge density is
A
$7.27 \times {10^{ - 4}}\,C/m$
B
$7.98 \times {10^{ - 4}}\,C/m$
C
$7.11 \times {10^{ - 4}}\,C/m$
D
$7.04 \times {10^{ - 4}}\,C/m$
Solution
(b) Relation for electric field is given by $E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}$
(Given : $E = 7.182 \times 10^8\, N/C$)
$r = 2\, cm = 2 \times 10^{-2}\, m$
$\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^{ – 9}}$ $==>$ $\lambda = 2\pi {\varepsilon _0}rE = \frac{{2 \times 2\pi {\varepsilon _0}rE}}{2}$
$ = \frac{{1 \times 2 \times {{10}^{ – 2}} \times 7.182 \times {{10}^8}}}{{2 \times 9 \times {{10}^9}}} = 7.98 \times {10^{ – 4}}\,C/m$
Standard 12
Physics