Question 1a and 1b. Show your work for full credits! Consider a capacitor create

Question

Question 1a and 1b. Show your work for full credits!

Consider a capacitor created when two identical conducting plates are placed parallel and close to each in other in vacuum. The surface area for each plate is 0.0400 nr. The two plates are separated by a distance of 0.00200 m. When an electric potential difference of 50.0 V is established between the two plates, what is the magnitude of the electrostatic force between the charged plates? Is the force attractive or repulsive? (Answer: F = 1.10x1 Or4 N; attractive)What mass on the Earth surface will expenence a gravitational force with the same magnitude as the above electrostatic force?

Explanation / Answer

2.

Electric fileld = V/d ..

V is the electric potential difference

d - is the distance between plates

Electric field = 50/0.002 = 25000 N/C

Force = Electric Field* charge

Charge = c*v = e*A*V/(2*d )= 1/(4*pi*9) *10^-9 * 0.04 * 25000/2

Force = charge * 25000 = 1/(4*pi*9) *10^-9 * 0.04 * 25000*25000/2 = 0.00011 N

= 1.1*10^-4 N

Mass = Force/g = 0.0000112 Kg = 11.2 mg

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