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Force of attraction between the plates o...

Force of attraction between the plates of a parallel plate capacitor is

A

`(q^(2))/(2epsilon_(0)AK)`

B

`(q^(2))/(epsilon_(0)AK)`

C

`(q)/(2epsilon_(0)A)`

D

`(q^(2))/(2epsilon_(0)A^(2)K)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the force of attraction between the plates of a parallel plate capacitor, we can follow these steps: ### Step 1: Understand the Electric Field between the Plates The electric field (E) between the plates of a parallel plate capacitor can be expressed as: \[ E = \frac{\sigma}{\epsilon_0} \] where: - \( \sigma \) is the surface charge density (charge per unit area), - \( \epsilon_0 \) is the permittivity of free space. If a dielectric material with dielectric constant \( k \) is inserted, the electric field becomes: \[ E = \frac{\sigma}{k \epsilon_0} \] ### Step 2: Relate Surface Charge Density to Charge The surface charge density \( \sigma \) can be expressed in terms of the total charge \( Q \) and the area \( A \) of the plates: \[ \sigma = \frac{Q}{A} \] ### Step 3: Substitute \( \sigma \) into the Electric Field Equation Substituting \( \sigma \) into the electric field equation gives: \[ E = \frac{Q}{A k \epsilon_0} \] ### Step 4: Calculate the Force on One Plate The force \( F \) on one plate due to the electric field created by the other plate can be calculated using the formula: \[ F = Q \cdot E \] Substituting the expression for \( E \): \[ F = Q \cdot \left(\frac{Q}{A k \epsilon_0}\right) \] This simplifies to: \[ F = \frac{Q^2}{A k \epsilon_0} \] ### Step 5: Final Expression for the Force Since the force is acting between the plates, we can express it as: \[ F = \frac{1}{2} \cdot \frac{Q^2}{A \epsilon_0 k} \] ### Conclusion Thus, the force of attraction between the plates of a parallel plate capacitor is given by: \[ F = \frac{Q^2}{2 A \epsilon_0 k} \]
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