A) 0.9 N
B) 1.8 N
C) 2.72 N
D) 3.12 N
Correct Answer: D
Solution :
\[{{F}_{AC}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{A}}{{q}_{C}}}{{{(AC)}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 1\times {{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(10\times {{10}^{-2}})}^{2}}}\] \[=1.8\] Similarly, \[{{F}_{BC}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}{{q}_{C}}}{{{(BC)}^{2}}}\] \[=\frac{9\times {{10}^{9}}\times 1\times {{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(10\times {{10}^{-2}})}^{2}}}\] \[=1.8\] \[{{F}_{resul\tan t}}=\sqrt{F_{AC}^{2}+F_{BC}^{2}+2{{F}_{AC}}{{F}_{BC}}\cos {{60}^{o}}}\] \[=\sqrt{{{(1.8)}^{2}}+{{(1.8)}^{2}}+2{{(1.8)}^{2}}\times \frac{1}{2}}\] \[=3.12\,N\]You need to login to perform this action.
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