Two identical metal plates are given positive charges Q1 and Q2 (

Question

Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,the potential difference between them is (A) (Q1+Q2)/2C  (B) (Q1+Q2)/C  (C) (Q1-Q2)/C  (D) (Q1-Q2)/2C

Tardigrade Admin · Accepted Answer

Electric field within the plates E = EQ1+EQ2 E=E1-E2= (Q1/2A ε0)- (Q2/2Aε0),E= (Q1-Q2/2Aε0) ∴ Potential difference between the plates VA-VB=ED= ( (Q1-Q2/2A ε0) )d= (Q1-Q2/2 ( (A ε0)d ))= (Q1-Q2/2C)

Q. Two identical metal plates are given positive charges $Q_{1}$ and $Q_{2} (< Q_{1})$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,the potential difference between them is

$(Q_{1} + Q_{2}) /2 C$

$(Q_{1} + Q_{2}) / C$

$(Q_{1} - Q_{2}) / C$

$(Q_{1} - Q_{2}) /2 C$

Q. Two identical metal plates are given positive charges Q1​ and Q2​(<Q1​) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,the potential difference between them is

(Q1​+Q2​)/2C

(Q1​+Q2​)/C

(Q1​−Q2​)/C

(Q1​−Q2​)/2C

Electric field within the plates E =EQ1​​+EQ2​​ E=E1​−E2​=2Aε0​Q1​​−2Aε0​Q2​​,E=2Aε0​Q1​−Q2​​ ∴ Potential difference between the plates VA​−VB​=ED=(2Aε0​Q1​−Q2​​)d=2(dAε0​​)Q1​−Q2​​=2CQ1​−Q2​​

Q. Two identical metal plates are given positive charges $Q_{1}$ and $Q_{2} (< Q_{1})$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C,the potential difference between them is

$(Q_{1} + Q_{2}) /2 C$

$(Q_{1} + Q_{2}) / C$

$(Q_{1} - Q_{2}) / C$

$(Q_{1} - Q_{2}) /2 C$