question_answer

Equal charges q each are placed at the vertices of an equilateral triangle of side r. The magnitude of electric field intensity at any vertex is

A) \[\frac{2q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]                  

B) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]

C) \[\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]    

D) \[\frac{\sqrt{2}q}{4\pi {{\varepsilon }_{0}}.{{r}^{2}}}\]

Correct Answer: C

Solution :

The magnitude of intensity of electric field at point C due to charge at A and B is                    \[56u=\frac{56}{14}Natom=4Natom\]                 So, the net intensity  at point C is              \[\begin{align}   & CaC{{O}_{3}}\xrightarrow{Heat}CaO+C{{O}_{2}}\uparrow  \\  & \text{XColourless} \\  & \text{gas} \\ \end{align}\]                    \[\begin{align}   & CaO+{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}} \\  & \text{ResidueY} \\ \end{align}\] \[\begin{align}   & Ca{{\left( OH \right)}_{20}}+2C{{O}_{2}}\to Ca{{\left( HC{{O}_{3}} \right)}_{2}} \\  & \text{YExcessZ} \\ \end{align}\]        \[\begin{align}   & Ca{{\left( HC{{O}_{3}} \right)}_{2}}\xrightarrow{Heat}CaC{{O}_{3}}+C{{O}_{2}}\uparrow +{{H}_{2}}O \\  & ZX \\ \end{align}\]