Question

The figure shows four pairs of parallel identical conducting plates, separated by the same distance of 2.0 cm and arranged perpendicular to the x-axis. The electric potential of each plate is mentioned

(i) For which pair of the plates is the electric field E alongˆ i?

• A. I
• B. II
• C. III
• D. IV

Hint:

When solving problems involving electric fields and potentials, it is crucial to consider the sign of charge carriers and the direction of the electric field to determine movement and potential changes correctly.

Answer 1

The Correct Option is D

The electric field between two parallel plates is given by:
\[ \vec{E} = \frac{V}{d} \hat{i} \] where \( V \) is the potential difference between the plates, \( d \) is the distance between the plates, and \( \hat{i} \) is the unit vector along the x-axis.
For the electric field to be along \( \hat{i} \), the potential difference between the plates must be such that the electric field is directed along the x-axis. Among the options, the electric field between plates IV has this configuration, as the potential difference is from \( -100 \, V \) to \( -400 \, V \), and the plates are perpendicular to the x-axis.
% Question 29(ii) (ii) An electron is released midway between the plates of pair IV. It will:
% Option (A) move along \( \hat{i} \) at constant speed
% Option (B) move along - \( \hat{i} \) at constant speed
% Option (C) accelerate along \( \hat{i} \)
% Option (D) accelerate along -\( \hat{i} \)
% Correct Answer % Correct Answer Correct Answer:} (D) accelerate along \( -\hat{i} \) % Solution % Solution Solution: The electric field between the plates of pair IV is non-zero, and since the electron is midway between the plates, it will experience a force due to the electric field. This force will cause the electron to accelerate along the direction of the electric field. The direction of the field is from the positively charged plate to the negatively charged plate, i.e., along the \( -\hat{i} \) direction. % Question 29(iii) (iii) Let \( V_0 \) be the potential at the left plate of any set, taken to be at \( x = 0 \, m} \). Then potential \( V \) at any point \( 0 \leq x \leq 2 \, cm} \) between the plates of that set can be expressed as: % Option % Option (A) \( V = V_0 + \alpha x \)
% Option % Option (B) \( V = V_0 + \alpha x^2 \)
% Option % Option (C) \( V = V_0 + \alpha x^{1/2} \)
% Option % Option (D) \( V = V_0 + \alpha x^{3/2} \) % Correct Answer % Correct Answer Correct Answer:} (A) \( V = V_0 + \alpha x \)
% Solution Solution: Assuming a linear variation of potential across the gap, the potential at any point \( x \) between the plates is a linear function of \( x \):
\[ V = V_0 + \alpha x \] where \( \alpha \) is the rate of change of potential per unit distance, determined by the difference in potential across the plates and the distance between them.
Answer: (A) \( V = V_0 + \alpha x \) is the correct expression for the potential \( V \) at any point \( x \) between the plates.