question_answer

See the diagram. Area of each plate is \[2.0{{m}^{2}}\]and  \[d=2\times {{10}^{-3}}m\]. A charge   of \[8.85\times {{10}^{-8}}C\]is given to Q. Then the potential 2d of Q becomes           

A)  \[13V\]

B)  \[10V\]                        

C)  \[6.67V\]

D)  \[8.825V\]

Correct Answer: C

Solution :

: The plates P and Q will form one capacitor and plates Q and R will form another capacitor. These two capacitors are connected in parallel as shown in the figure. Their effective capacitance is \[{{C}_{eq}}=\frac{{{\varepsilon }_{0}}A}{d}+\frac{{{\varepsilon }_{0}}A}{2d}=\frac{3}{2}\frac{{{\varepsilon }_{0}}A}{d}\] Potential of plate Q is \[V=\frac{q}{{{C}_{eq}}}=\frac{2qd}{3{{\varepsilon }_{0}}A}\] Here, \[A=2.0{{m}^{2}},d=2\times {{10}^{-3}}m\] \[q=8.85\times {{10}^{-8}}C,{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\] \[\therefore \] \[V=\frac{2\times 8.85\times {{10}^{-8}}C\times 2\times {{10}^{-3}}m}{3\times 8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\times 2{{m}^{2}}}\] \[\frac{20}{3}V=6.67V\]