See the diagram. Area of each plate is `2.0 m^(2)` and `d = 2 xx 10^(-3) m`. A charge of `8.85 xx 10^(-8) C` is given to Q. Then the potential of Q becomes

Options

A. 13 V
B. 10 V
C. 6.67 V
D. 8.825 V

Answer

C

Explanation

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) = (epsi_(0)A)/(d) + (epsi_(0)A)/(2d) = (3)/(2) (epsi_(0)A)/(d)`
Potential of plate Q is
`V= (q)/(C_(eq)) = (2qd)/(3 epsi_(0)A)`
Here, `A= 2.0 m^(2), d= 2 xx 10^(-3) m`
`q= 8.85 xx 10^(-8)C, epsi_(0) = 8.85 xx 10^(-12)C^(2)N^(-1)m^(-2)`
`:. V= (2 xx 8.85 xx 10^(-8) C xx 2 xx 10^(-3)m)/(3 xx 8.85 xx 10^(-12) C^(2)N^(-1)m^(-2) xx 2m^(2))`
`= (20)/(3)V = 6.67V`