When a potential difference of 170 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 24.0 nC/cm2. What is the spacing between the plates?

0.0627 μcm or 6.27 x cm

Explanation:

The charge (Q) on the plates can be expressed in terms of the surface charge density (σ) as follows;

Q = σA ---------------(i)

Where A is the overlap area between the plates

Also,

We know that capacitance (C) of a parallel plate capacitor is related to charge (Q) and potential difference (V) as follows;

Q = CV   ------------(ii)

Combining the two equations,

CV = σA    ----------------- (iii)

Also,

The capacitance (C) of a parallel plate capacitor is given by;

C = ε₀ x A / d  --------------------(iv)

Where ε₀ = electric constant = 8.85 x F/cm

d = distance (spacing) between the plates.

Substituting of C in equation(iv) into equation(iii) gives;

=> (ε₀ x A / d) x  V =  σ x A

A appears on both sides of the equation, so we can cancel it out.

=> ε₀ / d x  V =  σ

Making d the subject of the formula gives;

=> d = (ε₀ x  V) / σ    -------------------------(v)

Substitute the values of ε₀, V = 170V and σ = 24.0nC/ into the equation (v)

=> d = 8.85 x x 170 / (24.0 x )

=> d = 0.0627 μcm or 6.27 x cm

Therefore the spacing between the plates is d = 0.0627 μcm or 6.27 x cm

statement #3 on andre"s list is false.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form