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 \[10\text{ }cm\]side respectively. The resultant force on the charge at C is

A)  \[0.9\text{ }N\]         

B)  \[1.8\text{ }N\]

C)  \[2.72\text{ }N\]

D)  \[3.12\text{ }N\]

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