Question

In the given figure each plate of capacitance C has value of charge –

• A. CE
• B. $\frac{CER_{1}}{R_{2}–r}$
• C. $\frac{CER_{2}}{R_{2} + r}$
• D. $\frac{CER_{1}}{R_{1}–r}$

Answer 1

Answer: (C)

$\frac{CER_{2}}{R_{2} + r}$

Answer 2

I = P.D across R₂ = IR₂ = $\frac { \mathrm { ER } _ { 2 } } { \mathrm { r } + \mathrm { R } _ { 2 } }$

q = $\frac { \mathrm { CER } _ { 2 } } { \mathrm { r } + \mathrm { R } _ { 2 } }$