The capacitance of a capacitor is 10 μF. The distance between the plates is made half and area is doubled. The capacitance of the capacitor will now be:
i think its 20 im not sure tho
No. Capacitance is proportional to A/d.
A doubles and d is reduced by half.
Both changes have the effect of doubling the capacitance.
Try again.
What happens when you double something twice?
so its 40?
To determine the capacitance of a capacitor after changing its distance between the plates and area, we can use the formula:
C = ε0 * (A / d)
Where:
C = Capacitance
ε0 = Permittivity of free space (constant)
A = Area of the plates
d = Distance between the plates
In this case, the capacitance of the initial capacitor is given as 10 μF. Let's denote this as C1 = 10 μF. We need to find the capacitance of the capacitor after halving the distance between the plates (d2 = d/2) and doubling the area of the plates (A2 = 2A).
Substituting these values in the formula, we get:
C2 = ε0 * (A2 / d2)
Now, let's apply the changes in the distance and area:
A2 = 2A
d2 = d/2
Substituting these values into the equation, we get:
C2 = ε0 * (2A / (d/2))
Simplifying:
C2 = ε0 * (2A * (2/d))
We know that the product of the distance and area remains constant since A * d = constant.
So, 2A * (2/d) = 4A/d = 4 * (A/d) = 4C1
Therefore, the capacitance of the capacitor after the changes will be 4 times the initial capacitance:
C2 = 4 * C1 = 4 * 10 μF = 40 μF
Hence, the capacitance of the capacitor will now be 40 μF, not 20 μF.