A) \[4\times {{10}^{-11}}C\]
B) \[2\times {{10}^{-11}}C\]
C) \[5.4\times {{10}^{-11}}C\]
D) \[7.5\times {{10}^{-11}}C\]
Correct Answer: B
Solution :
The electric field E due to point charge q is given by \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\] \[\Rightarrow \] \[q=E\cdot 4\pi {{\varepsilon }_{0}}{{r}^{2}}\] Given,\[r=30\,\,cm=30\times {{10}^{-2}}m,\,\,E=2N/C\] \[\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\] Hence,\[q=2\times \frac{1}{9\times {{10}^{9}}}{{\left( \frac{30}{100} \right)}^{2}}=2\times {{10}^{-11}}C\]
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