Two identical thin metal plates has charge q1 and q 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 :

Question

Two identical thin metal plates has charge q1 and q 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 : (A) ((q1+q2)/C) (B) ((q1-q2)/C) (C) ((q1- q 2)/2 C ) (D) (2(q1-q2)/ C )

Tardigrade Admin · Accepted Answer

Electric field between plates E =( q 1- q 2/2 A ϵ0) V = Ed =( q 1- q 2/2 A ∈0) d V =( q 1- q 2/2 C )

Q. Two identical thin metal plates has charge $q_{1}$ and $q_{2}$ 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 :

$\frac{( q _{1} + q _{2} )}{C}$

$\frac{( q _{1} - q _{2} )}{C}$

$\frac{( q _{1} - q _{2} )}{2 C}$

$\frac{2 ( q _{1} - q _{2} )}{C}$

Electric field between plates $E = \frac{q _{1} - q _{2}}{2 A ϵ _{0}}$
$V = E d = \frac{q _{1} - q _{2}}{2 A \in _{0}} d$
$V = \frac{q _{1} - q _{2}}{2 C}$

Q. Two identical thin metal plates has charge q1​ and q2​ respectively such that q1​>q2​. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :

C(q1​+q2​)​

C(q1​−q2​)​

2C(q1​−q2​)​

C2(q1​−q2​)​