Verified Solution • 2025–26 Session moderate

QPages 165-170: Q22: HC Verma: Concepts of Physics - Volume 2 Capacitors Pages 165-170: Q22 Solution: Both the capacitors shown in figure are made of square plates…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse the direction of the electric field with the force on the electron, leading to incorrect sign conventions in the kinematic equations. Remember that the force on an electron is opposite to the electric field direction, so the acceleration term in the deflection calculation must account for this reversal.

Both the capacitors shown in figure are made of square plates of edge a. The separations between the plates of the capacitors are d1d_1 and d2d_2 as shown in the figure. A potential difference V is applied between the points a and b. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate? Consider only the electric forces.

Step-by-Step Verified Solution

**Answer:** v=eVm(d1d2)v = \sqrt{\frac{e V}{m}\left(\frac{d_1}{d_2}\right)}. **Step-by-Step Explanation:** The electron must not hit the plates, so its vertical deflection must be less than half the plate separation. Using kinematic equations and the electric field between the plates, the minimum speed is derived.

All Questions in this Chapter

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