A) \[\text{1}\text{.64 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-26}}}\,\text{N,2}\text{.4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-16}\,}}\text{N}\]
B) \[\text{1}\text{.64 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-26}}}\,\text{N,1}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}\,}}\text{N}\]
C) \[\text{1}\text{.56 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-18}}}\,\text{N,2}\text{.4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-16}}}\text{N}\]
D) \[\text{1}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\,\text{N,2}\text{.4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-16}}}\text{N}\]
Correct Answer: A
Solution :
For electrostatic force, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] \[3\times {{10}^{-6}}=\frac{9\times {{10}^{9}}\times 2\times {{10}^{-9}}\times 2\times {{10}^{-9}}}{{{r}^{2}}}\] \[\therefore {{r}^{2}}=\frac{9\times {{10}^{9}}\times 2\times {{10}^{-9}}\times 2\times {{10}^{-9}}}{3\times {{10}^{-6}}}\] Electrostatic force, when proton placed in this field, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\]\[\times \frac{Charge\text{ }of\text{ }partice\text{ }\times \text{ }charge\text{ }of\text{ }proton}{{{r}^{2}}}\] \[F=\frac{\frac{9\times {{10}^{9}}\times 2\times {{10}^{-9}}\times 1.6\times {{10}^{-19}}}{9\times {{10}^{9}}\times 2\times {{10}^{-9}}\times 2\times {{10}^{-9}}}}{3\times {{10}^{-6}}}\] \[F=2.4\times {{10}^{-16}}N\] Similarly, we can calculate gravitational force \[=1.64\times {{10}^{-26}}N\]
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