Question

The electric field between the plates of a parallel plate capacitor is 'E'. If the charge on the plates is Q then the force on each plate is

• A. \(\text{QE}^2\)
• B. \(\text{QE}\)
• C. \(\frac{\text{QE}}{2}\)
• D. \(\frac{\text{QE}^2}{2}\)

Hint:

The field of one plate is E/2, and it acts on the charge of the other plate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The field \(E\) between the plates is the sum of the fields from the two plates, each contributing \(E/2\).

Step 2: Force on one plate:
The charge \(Q\) on one plate feels only the field of the other plate, not its own. So
\[ F = Q\cdot\frac E2 = \frac{QE}{2} \]

Step 3: Check:
Option (C). Option (B) \(QE\) would use the total field, which includes the plate's own field that exerts no net force on itself. Options (A) and (D) have \(E^2\), which does not fit the unit of force here.

Final Answer:
The other plate gives E/2, so F = QE/2. \[ \boxed{\text{(C) }\dfrac{QE}{2}} \]