Question

9. A 1 nF parallel plate capacitor has plates with a charge of 1μC and -1μC. A particle with a charge of -4μC and a mass of 2x10-16 kg is fired from the positive plate toward the negative plate with an initial speed of 2x106 m/s. Does the particle reach the negative plate? What fraction of the distance does it cover between the plates before it turns around? ans(a=No,b=1/10)

Answer 1

C = Capacitance = 1 x 10^-9 F

Q = charge = 1 x 10^-6 C

V = potential difference = Q/C = 1 x 10^-6 /(1 x 10^-9 ) = 1000 volts

To reach the opposite plate , the change in electric potential energy must be

U = QV = 1 x 10^-6 (1000) = 0.001 J

kinetic energy is given as

KE = (0.5) m v² = (0.5) (2 x 10^-16) (2 x 10⁶)² = 0.0004 J

since KE < U

hence particle does not reach negative plate

let the distance between the plates = d

distance at which particle returns = x

E = electric field = 1000/d

U' = actual change in potential energy = QE x = 1000 Qx/d

using conservation of energy

U' = KE

1000 Qx/d = 0.0004

1000 (1 x 10^-6) (x/d ) = 0.0004

x/d = 0.4