Respuesta :
Explanation:
(A) Electric field for the parallel plate capacitor is given by :
[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]
It is clear that the electric field does not depend on the separation of the plates.
(B) The relation between the electric field and the electric potential is given by :
[tex]V=Ed[/tex]
d is the separation between plates. So, if the separation of the plates is increased, the potential difference increases.
(C) The capacitance of the parallel plate capacitor is given by :
[tex]C=\dfrac{A\epsilon_o}{d}[/tex]
So, the capacitance decreases when the separation of the plates is increased.
(D) The energy stored in the capacitor is given by :
[tex]E=\dfrac{1}{2}CV^2[/tex]
[tex]E=\dfrac{1}{2}C(Ed)^2[/tex]
So, the energy stored in the capacitor is increased when the separation of the plates is increased.
Answer:
a)constant
b)constant
c)constant
d) constant
Explanation:
a)
The electric field between the plates remain constant. The Electric field between the plates is given as:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
where:
[tex]\sigma=[/tex] surface charge density
[tex]\epsilon=[/tex] permittivity of the material between the plates
b)
The potential difference between the plates is related as:
[tex]V=\frac{Q}{C}[/tex]
and
[tex]E=\frac{V}{d}[/tex]
where:
d = distance between the plates
Therefore the potential difference remains constant when the capacitor plates distance remains constant.
c)
the capacitance:
[tex]C=\frac{Q}{V}[/tex]
When the charge and potential difference is constant then the capacitance also remains constant.
d)
Energy stored in a capacitor:
[tex]U=\frac{1}{2} C.V^2[/tex]
Since capacitance and potential difference are constant therefore potential difference is also constant.
