Question

An arrangement of source charges produces the electric potential V=5000x2V=5000x2 along the x-axis, where V is in volts and x is in meters. What is the maximum speed of a 1.0 g, 10 nC charged particle that moves in this potential with turning points at ± 8.0 cm?

Answer 1

Answer:

v = 0.025 m/s

Explanation:

Given that the voltage is

$V = 5000 x^2$

now at x = 0

$V_1 = 0 Volts$

also we have at x = 8 cm

$V_2 = 5000(0.08)^2 = 32 Volts$

now change in potential energy of the charge is given as

$\Delta U = q\Delta V$

$\Delta U = (10 \times 10^{-9})(32 - 0)$

now by mechanical energy conservation law

$\frac{1}{2}mv^2 - 0 = 3.2 \times 10^{-7}$

$\frac{1}{2}(1 \times 10^{-3})v^2 = 3.2 \times 10^{-7}$

$v = 0.025 m/s$