question_answer

Three charges \[1\mu C,\,1\mu C\]and \[2\mu C\] are kept at vertices of A, B and C of an equilateral triangle ABC of 10 cm side respectively. The resultant force on the charge at C is

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,                 \[{{F}_{BC}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}{{q}_{C}}}{{{(BC)}^{2}}}\]                 \[=\frac{9\times {{10}^{9}}\times 1\times {{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(10\times {{10}^{-2}})}^{2}}}\]                 \[=1.8\] \[{{F}_{resul\tan t}}=\sqrt{F_{AC}^{2}+F_{BC}^{2}+2{{F}_{AC}}{{F}_{BC}}\cos {{60}^{o}}}\]                 \[=\sqrt{{{(1.8)}^{2}}+{{(1.8)}^{2}}+2{{(1.8)}^{2}}\times \frac{1}{2}}\]                 \[=3.12\,N\]