Question

An electric potential of 4.93×103x2 V (where x is measured in cm) is acting along the x-axis. A particle with a charge of 6.20 nC and a mass of 7.80 g moves in this electric potential with turning points of ±2.90 cm. What is the particle's maximum speed through the electric potential?

A. 1.06 m/s

B 0.257 m/s

C -0.543 m/s

D 0.128 m/s

Answer 1

we have

(1/2)*(m)*((v_max)^2) = q*V;

Setting kinetic energy equal to each others

(1/2)*(7.80 grams)*((v_max)^2) = (6.20 nanocoulomb)*(4930*(x^2));

(1/2)*(7.80 grams)*((v_max)^2) = (6.20 nanocoulomb)*(4930 volts/cm^2)*((2.90 cm)^2)

(v_max)^2 = (6.20 nanocoulomb)*(4930 volts/cm^2)*((2.90 cm)^2)/[(1/2)*(7.80 grams)];

(v_max)^2 = (514120.12/7.80)*10^(-6) ((meter/second)^2);

v_max = sqrt(65912.8359*10^(-6)) meter/second

v_max= 256.73*10^(-3) m/s = 0.2567 m/s = 0.257 m/s

answer is part B