JEE Advanced PYQ Physics – Electrostatic Potential and Capacitance Mixed Concepts Questions

Q 1 :

Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference $X$ . A proton is released at rest midway between the two plates. It is found to move at $45 °$ to the vertical JUST after release. Then $X$ is nearly [2012]

$1 \times 10^{- 5} V$
$1 \times 10^{- 7} V$
$1 \times 10^{- 9} V$
$1 \times 10^{- 10} V$

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(3)

According to the question, a proton is released at rest midway between the two plates and is found to move at $45 °$ ,

Therefore, the net force is at $45 °$ from the vertical, and the two forces acting on the proton just after release are as shown in the figure.

$q E = m g$

$∴$ $q (\frac{V}{d}) = m g$

$∴$ $V = \frac{m g d}{q}$ $= \frac{1.67 \times 10^{- 27} \times 10 \times 10^{- 2}}{1.6 \times 10^{- 19}}$ $\approx 10^{- 9} V$

Hence, $X \approx 1 \times 10^{- 9} V$

Q 2 :

Consider a simple RC circuit as shown in Figure 1.

Process 1: In the circuit the switch S is closed at $t = 0$ and the capacitor is fully charged to the voltage $V_{0}$ (i.e., charging continues for time $T ≫ R C$ ). In the process some dissipation $(E_{D})$ occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is $E_{C}$ .

Process 2: In a different process the voltage is first set to $\frac{V_{0}}{3}$ and maintained for a charging time $T ≫ R C$ . Then the voltage is raised to $\frac{2 V_{0}}{3}$ without discharging the capacitor and again maintained for a time $T ≫ R C$ . The process is repeated one more time by raising the voltage to $V_{0}$ and the capacitor is charged to the same final voltage $V_{0}$ as in Process 1.

These two processes are depicted in Figure 2.

Q. In Process 1, the energy stored in the capacitor $E_{C}$ and heat dissipated across resistance $E_{D}$ are related by: [2017]

$E_{C} = E_{D}$
$E_{C} = E_{D} \ln 2$
$E_{C} = \frac{1}{2} E_{D}$
$E_{C} = 2 E_{D}$

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(1)

In process 1, work done by the battery, $W = q \times V$ $= C V_{0} \times V_{0}$ $= C V_{0}^{2}$

Energy stored in the battery, $E_{C} = \frac{1}{2} C V_{0}^{2}$

Heat dissipated across the resistance,

$E_{D} = W - E_{C}$ $= C V_{0}^{2} - \frac{1}{2} C V_{0}^{2}$ $= \frac{1}{2} C V_{0}^{2}$

Therefore, $E_{C} = E_{D}$

Q 3 :

Consider a simple RC circuit as shown in Figure 1.

Process 1: In the circuit the switch S is closed at $t = 0$ and the capacitor is fully charged to the voltage $V_{0}$ (i.e., charging continues for time $T ≫ R C$ ). In the process some dissipation $(E_{D})$ occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is $E_{C}$ .

Process 2: In a different process the voltage is first set to $\frac{V_{0}}{3}$ and maintained for a charging time $T ≫ R C$ . Then the voltage is raised to $\frac{2 V_{0}}{3}$ without discharging the capacitor and again maintained for a time $T ≫ R C$ . The process is repeated one more time by raising the voltage to $V_{0}$ and the capacitor is charged to the same final voltage $V_{0}$ as in Process 1.

These two processes are depicted in Figure 2.

Q. In Process 2, total energy dissipated across the resistance $E_{D}$ is: [2017]

$E_{D} = \frac{1}{2} C V_{0}^{2}$
$E_{D} = 3 (\frac{1}{2} C V_{0}^{2})$
$E_{D} = \frac{1}{3} (\frac{1}{2} C V_{0}^{2})$
$E_{D} = 3 C V_{0}^{2}$

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(3)

Let $V_{i}$ and $V_{f}$ be the initial and final voltages in each step of process 2. Then,

$Energy dissipated = W_{battery} - Δ U$

$= C (V_{f} - V_{i}) V_{f} - \frac{1}{2} C {(V_{f} - V_{i})}^{2}$

$= \frac{1}{2} C {(V_{f} - V_{i})}^{2}$

Therefore, the total energy dissipated across the resistance is

$E_{D} = \frac{1}{2} C [{(\frac{V_{0}}{3} - 0)}^{2} + {(\frac{2 V_{0}}{3} - \frac{V_{0}}{3})}^{2} + {(V_{0} - \frac{2 V_{0}}{3})}^{2}]$

$= \frac{1}{2} C [\frac{V_{0}^{2}}{9} + \frac{V_{0}^{2}}{9} + \frac{V_{0}^{2}}{9}]$

$= \frac{1}{2} C (\frac{V_{0}^{2}}{3})$

$= \frac{1}{6} C V_{0}^{2}$

Hence, $E_{D} = \frac{1}{6} C V_{0}^{2}$

Q 4 :

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius $r ≪ h$ . Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at $+ V_{0}$ and the top plate at $- V_{0}$ . Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. Ignore gravity.

Q. Which one of the following statements is correct? [2016]

The balls will stick to the top plate and remain there
The balls will bounce back to the bottom plate carrying the same charge they went up with
The balls will bounce back to the bottom plate carrying the opposite charge they went up with
The balls will execute simple harmonic motion between the two plates

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(3)

After colliding the top plate, the ball will gain negative charge and get repelled by the top plate and bounce back to the bottom plate. But ball do not execute simple harmonic motion as force on it $\propto x$

Q 5 :

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius $r ≪ h$ . Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at $+ V_{0}$ and the top plate at $- V_{0}$ . Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. Ignore gravity.

Q. The average current in the steady state registered by the ammeter in the circuit will be [2016]

zero
proportional to the potential $V_{0}$
proportional to $V_{0}^{1 / 2}$
proportional to $V_{0}^{2}$

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(4)

Average current, $I_{av} \propto \frac{Q}{t} \dots (i)$

Here, $Q \propto V_{0} \dots (i i)$

From $s = u t + \frac{1}{2} a t^{2}$

$h = \frac{1}{2} \frac{Q E}{m} t^{2} = \frac{1}{2} (\frac{Q \times 2 V_{0}}{m h}) t^{2}$ $(∵ a = \frac{F}{m} = \frac{q E}{m})$

$∴$ $t \propto \frac{1}{V_{0}} \dots (i i i)$ $[∵ Q \propto V_{0}]$

From equations (i), (ii) and (iii),

$I_{av} \propto \frac{V_{0}}{1 / V_{0}}$

$∴$ $I_{av} \propto V_{0}^{2}$

Topic Question Set

Q 1 :

Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference $X$ . A proton is released at rest midway between the two plates. It is found to move at $45 °$ to the vertical JUST after release. Then $X$ is nearly [2012]

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Topic Question Set

Q 1 : Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45° to the vertical JUST after release. Then X is nearly [2012]

Want Us to Email you About Special Offers & Updates?

Q 1 :

Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference $X$ . A proton is released at rest midway between the two plates. It is found to move at $45 °$ to the vertical JUST after release. Then $X$ is nearly [2012]