Question

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

• A. $QE$
• B. $\frac{QE^2}{2}$
• C. $QE^2$
• D. $\frac{QE}{2}$

Hint:

To avoid a common pitfall, remember that writing $F = QE$ incorrectly implies that a plate experiences a force from its own electric field. Since the field from one plate is exactly half of the total field ($E_{\text{external}} = \frac{E}{2}$), the force must always be $\frac{QE}{2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem presents a parallel plate capacitor containing total internal electric field $E$ and plate charges $+Q$ and $-Q$. We need to find the attractive mechanical force exerted between the two oppositely charged plates.

Step 2: Key Formula or Approach:
The net internal field $E$ is the vector sum of the independent electric fields produced by both individual plates: $$E = E_{\text{plate 1}} + E_{\text{plate 2}}$$ Since a plate cannot exert a force on itself, the force acting on one plate is driven exclusively by the electric field generated by the other plate. We calculate this force using: $$F = Q \times E_{\text{other}}$$

Step 3: Detailed Explanation:
The total field between the two plates of a parallel-plate capacitor is given by: $$E = \frac{\sigma}{\varepsilon_0}$$ Because both plates are identical, they contribute equally to this field. The field produced by a single plate is: $$E_{\text{single}} = \frac{\sigma}{2\varepsilon_0} = \frac{E}{2}$$ The force acting on the second plate carrying charge $Q$ due to this field is: $$F = Q \cdot E_{\text{single}} = Q \cdot \left(\frac{E}{2}\right) = \frac{QE}{2}$$

Step 4: Final Answer:
The attractive force acting on each plate is $\frac{QE}{2}$, which corresponds to option (D).