An experimental design is used to assign variables for testing. In contrast to a control design where nothing is changed, the experimental design allows you to test various new inputs to see how they would vary from the original results.
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.
Answer:
2.8 cm
Explanation:
= Separation between two first order diffraction minima = 1.4 cm
D = Distance of screen = 1.2 m
m = Order
Fringe width is given by

Fringe width is also given by

For second order

Distance between two second order minima is given by


The distance between the two second order minima is 2.8 cm
Answer:
when a element of 1 group take part in reaction, its atom looses outer electron and form positively charged ions called Cation.
Explanation: