A) \[-7.5\times {{10}^{-5}}C\]
B) \[-7.5\times {{10}^{-9}}C\]
C) \[-7.5\times {{10}^{-5}}C\]
D) \[-7.5\times {{10}^{-9}}C\]
Correct Answer: B
Solution :
Electric field, \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{r}^{2}}}\] \[3\times {{10}^{3}}=9\times {{10}^{9}}\times \frac{q}{{{(15)}^{2}}\times {{10}^{-4}}}\] \[\therefore \] \[q=\frac{3\times {{10}^{3}}\times 15\times 15\times {{10}^{-4}}}{9\times {{10}^{9}}}\] \[q=75\times {{10}^{-10}}\] \[=7.5\times {{10}^{-9}}N/C\] \[q=-7.5\times {{10}^{-9}}N/C\] Charge will be negative because direction of electric field towards the charge.
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