A parallel plate capacitor has a charge on one plate of q = 4.5E-07 C. Each square plate is d1 = 1.7 cm wide and the plates of the capacitor are separated by d2 = 0.35 mm. The gap is filled with air, εo = 8.85 × 10-12 C2/Nm2.

Part (a) What is the voltage between the plates, Δ V, in V?

Part (b) What plate width would double this voltage, in centimeters?

Explanation:

capacitance of capacitor

= ε₀ A / d , A is area of plate , d is plate separation

= (8.85 x10⁻¹² x 1.7² x 10⁻⁴) /( .35 x 10⁻³ )

= 73 x 10⁻¹³ F

Charge = 4.5 x 10⁻⁷

voltage between the plates,

Δ V = charge / capacitance

= 4.5 x 10⁻⁷ / 73 x 10⁻¹³

= .0616 x 10⁶

= 616 x 10² V

To make the vottage double , we will reduce capacitance to half .

To reduce capacitance half we will increase plate distance to double.

new plate distance

= 1.7 x 2

= 3.4 cm

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2