Parallel Plate Capacitor Design Condition – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

Assertion (A): In parallel plate capacitor separation 'd' should be smaller than the linear dimension of the plates \( d^2 << A\). Reason (R): For \( d^2 << A \) a fringing effect can be ignored in the region sufficiently far from the edge.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution Explained:

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

Assertion (A) is true; for a parallel plate capacitor, 'd' must be much smaller than plate dimensions for the uniform field approximation. Reason (R) is also true. The condition \( d^2 << A \) (implying \( d << \sqrt{A} )\) allows ignoring fringing effects. Thus, (R) is the correct explanation for (A).

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A parallel plate capacitor is charged by a battery and after charging the capacitor, battery is disconnected and decrease the distance between the plates then which following statement is correct ?
2. A resistor 'R' and \(2~\mu\text{F}\) capacitor in series is connected through a switch to \(200~\text{V}\) direct supply. Across the capacitor is a neon bulb that lights up at \(120~\text{V}\). Calculate the value of R to make the bulb light up \(5~\text{s}\) after the switch has been closed. \((log_{10} 2.5 = 0.4)\)
3. A parallel plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their separation is 2d. What is the work required to separated the plates?
4. A capacitor of capacitance \(C\) is charged to a potential \(V\). The flux of the electric field through a closed surface enclosing the capacitor is
5. Two capacitors of capacitances \( C_1 \) and \( C_2 \) are connected in series across \( 200\text{V} \) power supply. The potential drop across \( C_1 \) is \( 120\text{V} \). A capacitor of capacitance \( 2\ \mu\text{F} \) is connected in parallel with \( C_1 \) and the potential drop across \( C_2 \) becomes \( 160\text{V} \). What are the values of \( C_1 \) and \( C_2 \) in \( \mu\text{F} \)?