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QJEEM22Phc909: JEE Main 2022 Physics - Class 12 JEEM22Phc909 Solution: Two identical thin metal plates has charge and respectively…

Two identical thin metal plates has charge q1q_{1} and q2q_{2} respectively such that q_{1}>q_{2} . The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

Multiple Choice Options

  • A (q1+q2)C\frac{\left(q_{1}+q_{2}\right)}{C}
  • B (q1−q2)C\frac{\left(q_{1}-q_{2}\right)}{C}
  • C (q1−q2)2C\frac{\left(q_{1}-q_{2}\right)}{2 C}
  • D 2(q1−q2)C\frac{2\left(q_{1}-q_{2}\right)}{C}

Step-by-Step Verified Solution

JEE Main 2022 (Online) 29th July Evening Shift Physics - Capacitor Question 44 English Explanation

Charge on the left surface of plate A=Total charge2\mathrm{A} = {{\mathrm{Total\,charge}} \over 2}

=q1+q22= {{{q_1} + {q_2}} \over 2}

Let right surface of plate A has charge =x= x

And total charge on plate A=q1A = {q_1}

∴\therefore q1q_1 = Charge on left surface of plate A + Charge on right surface of plate A

=q1+q22+x= {{{q_1} + {q_2}} \over 2} + x

⇒x=q1−q1+q22\Rightarrow x = {q_1} - {{{q_1} + {q_2}} \over 2}

=2q1−q1−q22= {{2{q_1} - {q_1} - {q_2}} \over 2}

=q1−q22= {{{q_1} - {q_2}} \over 2}

Let potential difference between two plates =V= V

For capacitor we know,

q=CVq = CV

∴\therefore q1−q22=CV{{{q_1} - {q_2}} \over 2} = CV

⇒V=q1−q22C\Rightarrow V = {{{q_1} - {q_2}} \over {2C}}

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