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QPages 121-124: Q64: HC Verma: Concepts of Physics - Volume 2 Electric Field Potential Pages 121-124: Q64 Solution: An electric field of magnitude is produced between two parallel …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the electric force on an electron as $F=+eE$ and use $v=\sqrt{2a d}$, forgetting that the electron’s charge is negative so the force (and acceleration) opposes the field direction. Use energy conservation $\frac12 m v^2 = e\,V$ (with $e=|e|$) to find the minimum speed, not a kinematic equation with a positive acceleration.

An electric field of magnitude

1000 \text{ N C}^{-1}

is produced between two parallel plates with a separation of 2.0 cm, as shown in the figure. a) What is the potential difference between the plates? b) With what minimum speed should an electron be projected from the lower plate in the direction of the field, so that it may reach the upper plate? c) Suppose the electron is projected from the lower plate with the speed calculated in part b). The direction of projection makes an angle of 60� with the field. Find the maximum height reached by the electron.

Step-by-Step Verified Solution

**Answer:** (a)

V = 20

V (b)

v = 8.4 \times 10^7

m/s (c)

s = 0.5

cm. **STEP-BY-STEP Explanation:** Given:

E = 1000

N/C,

d = 0.02

m. (a)

V = Ed = 1000 \times 0.02 = 20

V. (b)

a = eE/m = 1.76 \times 10^{14}

m/s_.

v = \sqrt{2ad} = \sqrt{2 \times 1.76 \times 10^{14} \times 0.02} = 8.4 \times 10^7

m/s. (c) For projection at 60�, maximum height

s = \frac{v^2\cos^2 60�}{2a} = 0.5

cm .

All Questions in this Chapter

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