The separation between the plates of an isola | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For an isolated capacitor,the charge $Q$ remains constant because there is no path for the charge to flow.The capacitance of a parallel plate capa

Vedclass · Accepted Answer

$B$ and $C$ both

Vedclass · Answer

(D) For an isolated capacitor,the charge $Q$ remains constant because there is no path for the charge to flow.The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$. As the separation $d$ increases,the capacitance $C$ decreases.The potential difference $V$ is given by $V = \frac{Q}{C}$. Since $Q$ is constant and $C$ decreases,the potential difference $V$ increases.The energy stored in the capacitor is given by $U = \frac{Q^2}{2C}$. Since $Q$ is constant and $C$ decreases,the energy $U$ increases.Therefore,both the potential difference and the energy of the capacitor change. The correct option is $D$.

The separation between the plates of an isolated charged parallel plate capacitor is increased. Which of the following quantities will change?

Similar Questions

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by the above two particles will be:

$A$ circuit shown in the figure consists of a battery of $emf$ $10\,V$ and two capacitors $C_1$ and $C_2$ of capacitances $1.0\,\mu F$ and $2.0\,\mu F$ respectively. The potential difference $V_A - V_B$ is $5\,V$. Which of the following statements is correct?

In the steady state,what is the potential difference between points $A$ and $B$ in volts? (All capacitors are in $\mu F$.)

If the potential of $A$ is $10\,V$,then the potential of $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module