Q23. Two identical thin metal plates are given charges q1 and q2 (q2 q1) respectively. If they are n

Q: What is the correct answer for CUET UG Physics Q23?

The correct answer is C. When two identical thin metal plates with charges q₁ and q₂ are brought close together to form a parallel plate capacitor, the charges redistribute themselves. The total charge (q₁ + q₂) will be distributed on the outer surfaces of the plates, with (q₁ + q₂)/2 on each outermost surface. The charges on the inner surfaces (facing each other) will be equal in magnitude and opposite in sign, given by

Q: Q23. Two identical thin metal plates are given charges q1 and q2 (q2 q1) respectively. If they are now brought close together to form a parallel plate capacitor with a capacitance 'C' then the…

The correct answer is C. When two identical thin metal plates with charges q₁ and q₂ are brought close together to form a parallel plate capacitor, the charges redistribute themselves. The total charge (q₁ + q₂) will be distributed on the outer surfaces of the plates, with (q₁ + q₂)/2 on each outermost surface. The charges on the inner surfaces (facing each other) will be equal in magnitude and opposite in sign, given by

Q: What are the options for this CUET UG Physics question?

A. (q1+q2)/2C · B. q1/2Cq2 · C. (q1-q2)/2C · D. q2/(C(q1+q2))

Question

Q23. Two identical thin metal plates are given charges q1 and q2 (q2 q1) respectively. If they are now brought close together to form a parallel plate capacitor with a capacitance 'C' then the potential difference between the plates is:
(a) (q1+q2)/2C
(b) q1/2Cq2
(c) (q1-q2)/2C
(d) q2/C(q1+q2)

FirstInTest · Accepted Answer

When two identical thin metal plates with charges q₁ and q₂ are brought close together to form a parallel plate capacitor, the charges redistribute themselves. The total charge (q₁ + q₂) will be distributed on the outer surfaces of the plates, with (q₁ + q₂)/2 on each outermost surface. The charges on the inner surfaces (facing each other) will be equal in magnitude and opposite in sign, given by Q_inner = (q₁ - q₂)/2 and -Q_inner = -(q₁ - q₂)/2. The potential difference (V) between the plates of a capacitor is directly related to the charge on its plates (Q_inner) and its capacitance (C) by V = Q_inner / C. Therefore, substituting the value of Q_inner, the potential difference between the plates is V = (q₁ - q₂)/(2C). This matches option (c).

Answer

(q1+q2)/2C

Answer

q1/2Cq2

Answer

(q1-q2)/2C

Answer

q2/(C(q1+q2))

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