Asked in MDCAT Test Series — Electrostatics, Work and EnergyModerate

An electron is situated midway between two parallel plates 0.5 cm apart. One of the plates is maintained at a potential of 60 volts above the other. The force on the electron is: (e = -1.6 x 10^-19C)

Correct answer: C. 1.92 x 10^-15 N

  • A. 1.82 x 10^-15 N
  • B. 3.00 x 10^-16 N
  • C. 1.92 x 10^-15 N
  • D. 3.00 x 10^-15 N
  • E. 5.00 x 10^-15 N

Explanation

To solve this problem, we first calculate the electric field (E) between the parallel plates using the formula E = V/d, where V is the potential difference and d is the distance between the plates. Here, V = 60V and d = 0.5 cm = 0.005 m. Thus, E = 60V/0.005m = 12,000 V/m.Next, we apply the formula for the force on a charge in an electric field: F = eE. The charge of an electron is e = -1.6x10-19 C. Therefore, F = (1.6x10-19 C)(12,000 V/m) = 1.92x10-15 N.Option A is incorrect due to a possible miscalculation of E. Option B underestimates the force, likely from an error in applying the formula. Option D overestimates the force, possibly by using incorrect values. Option E significantly overestimates the force, indicating a misunderstanding of the relationship between V, d, and the resulting force.

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About Electric Field Intensity

Electric field intensity is the force acting per unit positive test charge, with direction set by the force on that charge. Work includes fields of point charges, vector superposition, field lines and the relation between field and electric potential. Field intensity must be distinguished from electric force and potential.

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