Magnitude of a point charge due to which electric field 30,cm away has magnitude 2,newton/coulomb [1

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1. Electric Charges and Fields

easy

What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,newton/coulomb$ $[1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N{m^2}/{C^2}]$

$2 \times {10^{ - 11}}\,coulomb$

$3 \times {10^{ - 11}}\,coulomb$

$5 \times {10^{ - 11}}\,coulomb$

$9 \times {10^{ - 11}}\,coulomb$

Solution

(a) Electric field due to a point charge $E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}$
$q = E \times 4\pi {\varepsilon _0}{r^2} = 2 \times \frac{1}{{9 \times {{10}^9}}} \times {\left( {\frac{{30}}{{100}}} \right)^2}$= $2 \times {10^{ – 11}}\,coulomb$

Standard 12

Physics