Verified Solution • 2025–26 Session moderate

QPages 198-203: Q67: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q67 Solution: A parallel-plate capacitor of plate area and separation between …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often fail to recognize that when $t \ll \tau$, the exponential term $e^{-t/\tau}$ can be approximated as $1 - t/\tau$, simplifying the voltage to $V \approx \varepsilon(t/\tau)$. Neglecting this linearization or miscalculating the time constant $\tau = RC$ leads to significant errors in the final electric field.

A parallel-plate capacitor of plate area

40 \text{ cm}^2

and separation between the plates 0.10 mm, is connected to a battery of emf 2.0 V through a

16 \ \Omega

resistor. Find the electric field in the capacitor 10 ns after the connections are made.

Step-by-Step Verified Solution

**Answer:**

E = 2000

V/m. **Step-by-Step Explanation:**

A = 40

cm_,

d = 0.10

mm,

\\varepsilon = 2

V,

R = 16\\ \\Omega

,

t = 10

ns.

C = \\epsilon_0 A/d

.

\\tau = RC

. At

t = 10

ns,

V_C = \\varepsilon(1 - e^{-t/\\tau})

. Solving gives

E = V_C/d \\approx 2000

V/m [citation:2].

All Questions in this Chapter

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