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两块面积较大的金属薄板间距

d

较小，面积均为

S

、金属薄板分别有电荷量

Q

和

q

，则两者内部电荷间的相互作用力为（）

Mar 19, 2024

两块面积较大的金属薄板间距

d

较小，面积均为

S

、金属薄板分别有电荷量

Q

和

q

，则两者内部电荷间的相互作用力为（）

Solution

Identify the relevant concept: the force between two charged plates is given by Coulomb's law for parallel plate capacitors

Use the formula for the electric field between the plates:

E = \frac{\sigma}{\varepsilon_0}

where

\sigma = \frac{Q}{S}

is the surface charge density and

\varepsilon_0

is the vacuum permittivity

Calculate the force on the second plate due to the electric field created by the first plate:

F = qE

where

E

is the electric field and

q

is the charge on the second plate

Substitute the expression for

E

from step b into the equation from step c to find the force:

F = q \left(\frac{\sigma}{\varepsilon_0}\right) = q \left(\frac{Q}{S\varepsilon_0}\right)

Simplify to get the final expression for the force:

F = \frac{qQ}{S\varepsilon_0}

Answer

F = \frac{qQ}{S\varepsilon_0}

Key Concept

Coulomb's law for parallel plate capacitors and the concept of electric field and force between charged plates.

Explanation

The force between two charged plates is calculated using the electric field created by one plate and the charge on the other. The electric field is uniform between the plates and is directly proportional to the surface charge density and inversely proportional to the vacuum permittivity. The force on the second plate is then the product of its charge and the electric field.