Question

a parallel-plate capacitor is used to accelerate a positively charged particle, of mass 1.0x 10-3g and charge 1.0x10-10C, from one plate to another. The charge on each plate has a magnitude 1.0x10-5C, and the separation between the plates is 1.0x10-2m. The capacitor has a capacitance of 1.0x10-9F. what is the acceleration of the particle as it moves from one plate to the other?

Answer 1

Let the mass of the charged particle m = 1.0 × 10^-3 g

= 1.0 × 10^-6 kg

Charge of the particle q = 1.0 × 10^-10 C

Charge on the capacitor plate Q = 1.0 × 10^-5 C

Separation between the plates is d = 1.0 × 10^-2 m

Capacitance of the capacitor C = 1.0 × 10^-9 F

We need the electric field E between the plates to calculate the acceleration of the particle.

For this we first need potential difference between the plates.

Charge on a capacitor plate Q = C V

V = Q / C

= ( 1.0 × 10^-5 / 1.0 × 10^-9 )

= 10⁴ volt

Electric field E is given by E = potential difference / seperation

E = V / d

= 10⁴ / ( 1.0 × 10^-2 )

= 10⁶ V/m

Acceleration of the particle a = q E / m

= ( 10^-10 × 10⁶ / 10^-6 )

= 10² m/s²

= 100 m/s²

Acceleration of the particle as it moves from one plate to the other a = 100 m/s²