Respuesta :
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
The potential difference(voltage) of the plates if the plate spacing is doubled without changing Q is 800 V.
Given to us
Potential difference initially, ΔV = 400V
The initial distance between the plate, [tex]\rm d = d[/tex]
Final distance between the plate, [tex]\rm d_{new} = 2d[/tex] = 2d
What is the potential difference if the spacing of the plates is now doubled?
We know the capacitance of two parallel plates can be given as,
[tex]C=\dfrac{K\ \epsilon_0\ A }{d} = \dfrac{Q}{V}[/tex]
where C is the capacitance, k is the dielectric constant of the material, A is the are of the plates, d is the distance between the plates, Q is the charge between them, V is the potential difference.
On further simplification we get,
[tex]V = \dfrac{Qd}{K\ \epsilon_0\ A }[/tex]
Potential difference in the initial condition
In the initial condition substituting the values,
[tex]400 = \dfrac{Qd}{K\ \epsilon_0\ A }[/tex]
Potential difference if the space is double
In the final condition substituting the values we get,
[tex]V_{new} = \dfrac{Qd_{new}}{K\ \epsilon_0\ A }[/tex]
[tex]V_{new} = \dfrac{Q2d}{K\ \epsilon_0\ A }[/tex]
[tex]V_{new} = 2\dfrac{Qd}{K\ \epsilon_0\ A }[/tex]
[tex]V_{new} = 2 \times 400 = 800\rm\ V[/tex]
Hence, the potential difference(voltage) of the plates if the plate spacing is doubled without changing Q is 800 V.
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