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.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
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.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Electric Field Intensity questions
Electric field strength at a point between oppositely charged plates is 'E'. If the distance between plates is reduced to half, what will be the new value of electric intensity?
Electric field at a distance of 20cm from a 4µC charge is:
The electric field at a distance of 20cm from 4µC charge is:
SI unit of electric field is volt/meter, which is equivalent to:
The negative of the potential gradient is equal to:
What is the acceleration of an object having charge 2μC and mass 2 g moving through electric field strength 20 N/C?