Tag: capacitance

Questions Related to capacitance

A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^12V/ms.$ what is the displacement current?

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. 0.07$A$

  2. 1.39$A$

  3. 13.9$\mathrm { A }$

  4. 139$\mathrm { A }$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} Id={ \in _{ 0 } }\dfrac { { d\phi  } }{ { dt } } ={ \in _{ 0 } }\dfrac { { d\left( { EA } \right)  } }{ { dt } } ={ \in _{ 0 } }A\dfrac { { EA } }{ { dt } } =\in \dfrac { { \pi { r^{ 2 } }dE } }{ { dt } }  \ =8.85\times { 10^{ -12 } }\times 3.14\times 25\times { 10^{ -4 } }\times { 10^{ 12 } } \ =0.07A \end{array}$

Hence,
option $(A)$ is correct answer.

A capacitor is charged by a cell of emf $E$ and the charging battery is then removed. If an identical capacitor is now inserted in the circuit in parallel with the previous capacitor, the potential difference across the new capacitor is :

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. $2E$

  2. $E$

  3. $E/2$

  4. zero

Show answer
Correct Option: C
Explanation:

As the battery is disconnected so total is constant. i.e $Q _t=CE$
When a identical capacitor is add in parallel so the total capacitance is $C _t=C+C=2C$.
Now the common potential $\displaystyle =\frac{total \  charge }{total \  capacity}=\frac{CE}{2C}=\frac{E}{2}$