2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 30

  • question_answer
    Two parallel plate capacitors of equal plate area are connected as shown in the figure. Initially each capacitor is charged to charge Q and separation between the plates is d. One plate of each capacitor moves with constant speed v keeping the other plate of capacitor remain fixed, as indicated in the figure. The current developed in the circuit is

    A) \[\frac{Qv}{2d}\]                   

    B)        \[\frac{Qv}{d}\]                   

    C) \[\frac{3Qv}{d}\]                 

    D)        \[\frac{2Qv}{d}\]

    Correct Answer: B

    Solution :

    [b] \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d-vt},\,\,{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d+vt}\] \[\because \]  Potential difference across two capacitors is always equal. \[\Rightarrow \ \,\,\,\,\frac{q}{{{C}_{1}}}=\frac{(2Q-q)}{{{C}_{2}}}\] \[\Rightarrow \ \,\,\,\,\frac{q(d-vt)}{{{\varepsilon }_{0}}A}=\frac{(2Q-q)\,(d+vt)}{{{\varepsilon }_{0}}A}\] \[\Rightarrow \ \,\,\,\,q=\frac{Q}{d}(d+vt)\] \[\Rightarrow \ \,\,\,\,\frac{dq}{dt}=i=\frac{Qv}{d}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner