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Two plates with area A are held a distance d apart and have a net charge +Q, and -Q, respectively. Assume that all the charge is uniformly distributed on the inner surfaces of the plates.



The left plate has charge -Q, the right plate has charge +Q, separated by distance d.

1) Find the charge density on the plates.
2) Find the electric potential difference between the plates.
3) Show that the capacitance of the enlarged plates in this case is the same as the capacitance in a case where

Respuesta :

Answer:

Explanation:

1 )

Charge density of left plate

= - Q / A

Charge density of right plate

= + Q / A

2 )

capacitance c = ε₀ A / d

potential difference = charge / capacitance

= Q / [ ε₀ A / d ]

= Q d  / ε₀ A

(a) The charge density of the plates is [tex]\sigma=\frac{Q}{A}[/tex]

(b) The electric potential difference between the plates is [tex]V=\frac{Qd}{\epsilon_o A}[/tex]

(c) The capacitance between the plates is [tex]C=\frac{\epsilon_oA}{d}[/tex]

Capacitor:

Two plates with area A separated by a distance d and having charges +Q and -Q uniformly distributed form a capacitor.

(a)The charge per unit area is defined as the surface charge density. So the charge density on the plates is:

[tex]\sigma=\frac{Q}{A}[/tex]

(b) The electric potential difference between the plates in a capacitor is given by:

[tex]V=\frac{Q}{C}[/tex]

where C is the capacitance.

[tex]C=\frac{\epsilon_oA}{d}[/tex]

so the potential difference is:

[tex]V=\frac{Qd}{\epsilon_o A}[/tex]

(c) The capacitance of the plates is [tex]C=\frac{\epsilon_oA}{d}[/tex] as discussed above.

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