What is the magnitude of a point charge due to which the electric field $$30\,cm$$ away has the magnitude $$2\,newton/coulomb$$ $$[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}]$$ [MP PMT 1996]

Question

What is the magnitude of a point charge due to which the electric field $$30\,cm$$ away has the magnitude $$2\,newton/coulomb$$ $$[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}]$$                                                                                                       [MP PMT 1996]

TuteeHUB · Accepted Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	What is the magnitude of a point charge due to which the electric field \[30\,cm\] away has the magnitude \[2\,newton/coulomb\] \[[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}]\] [MP PMT 1996]
A.	\[2\times {{10}^{-11}}coulomb\]
B.	\[3\times {{10}^{-11}}coulomb\]
C.	\[5\times {{10}^{-11}}coulomb\]
D.	\[9\times {{10}^{-11}}coulomb\]
Answer» B. \[3\times {{10}^{-11}}coulomb\]

What is the magnitude of a point charge due to which the electric field \[30\,cm\] away has the magnitude \[2\,newton/coulomb\] \[[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}]\] [MP PMT 1996]

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