Q145 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
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$\text{QE}$
- Correct answerC$\dfrac{\text{QE}}{2}$
- D$\dfrac{\text{QE}^2}{2}$
Solution
The electric field between the plates of a parallel plate capacitor is the superposition of the electric fields produced by each plate.
The total electric field is $E = \dfrac{Q}{A\epsilon_0}$.
The electric field produced by a single plate is $E' = \dfrac{Q}{2A\epsilon_0} = \dfrac{E}{2}$.
The force experienced by one plate is due to the electric field of the other plate.
Force $F = Q \times E' = Q \times \dfrac{E}{2} = \dfrac{QE}{2}$.
Answer: $\dfrac{\text{QE}}{2}$
The total electric field is $E = \dfrac{Q}{A\epsilon_0}$.
The electric field produced by a single plate is $E' = \dfrac{Q}{2A\epsilon_0} = \dfrac{E}{2}$.
The force experienced by one plate is due to the electric field of the other plate.
Force $F = Q \times E' = Q \times \dfrac{E}{2} = \dfrac{QE}{2}$.
Answer: $\dfrac{\text{QE}}{2}$
Hence, the correct answer is option: 'C'.