Question

Figures show four pairs of parallel plates \(P,Q,R\) and \(S\) with the same separation and the electric potential of each plate. The electric field between the plates is uniform and perpendicular to the plates. Arrange the plates in descending order of the magnitude of the electric field between the plates.

(A) P (B) Q (C) R (D) S Choose the correct answer from the options given below:

• A. \((B),(C),(D),(A)\)
• B. \((A),(C),(B),(D)\)
• C. \((B),(C),(A),(D)\)
• D. \((C),(D),(A),(B)\)

Hint:

For parallel plates: \[ E=\frac{\Delta V}{d} \] If the plate separation is the same, simply compare the magnitudes of the potential differences. Larger \(|\Delta V|\) implies a stronger electric field.

Answer 1

The Correct Option is C

Concept: For parallel plates having the same separation \(d\), \[ E=\frac{\Delta V}{d} \] Therefore, the magnitude of electric field is directly proportional to the potential difference between the plates.

Step 1: Calculate the potential difference for each pair. For \(P\), \[ \Delta V_P = |100-(-50)| = 150\text{ V} \] For \(Q\), \[ \Delta V_Q = |200-(-20)| = 220\text{ V} \] For \(R\), \[ \Delta V_R = |-400-(-200)| = 200\text{ V} \] For \(S\), \[ \Delta V_S = |160-40| = 120\text{ V} \]

Step 2: Compare electric fields. Since all separations are equal, \[ E \propto \Delta V \] Thus, \[ 220>200>150>120 \] or \[ E_Q>E_R>E_P>E_S \]

Step 3: Write the descending order. \[ Q>R>P>S \] which corresponds to \[ (B),(C),(A),(D) \]

Step 4: State the answer. \[ \boxed{ Q>R>P>S } \] Hence, the correct option is \[ \boxed{(C)} \]