Answer:
Electric field will remain same.
Explanation:
Given that
At initial condition ,charge q = +3μ C
Electric field E = 4106 N/c
As we know that
Electric field due to charge q
Now when charge is replaced by new charge q'= -3μ C
From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.
So we can say that electric field will remain same.
(a) 12.0 V
In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, , the charged stored on the capacitor at the end of the process is
When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship
and since neither Q nor C change, the voltage V remains the same, 12.0 V.
(b) (i) 24.0 V
In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:
where:
is the permittivity of free space
is the relative permittivity of the material inside the capacitor
A is the area of the plates
d is the separation between the plates
As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: . The new voltage across the plate is given by
and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is
So, the new voltage is
(c) (ii) 3.0 V
The area of each plate of the capacitor is given by:
where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be
While the separation between the plate was unchanged (d); so, the new capacitance will be
So, the capacitance has increased by a factor 4; therefore, the new voltage is
which means