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 force on the thing and the distance the force moves the thing

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.