Question

Four point charges are fixed at the corners of a square centered at the origin. The length of each side of the square is 2a. The charges as located as follows: +q is at (-a, +a), +3q is at (+a, +a), -4q is at (+a, -a), and +10q is at (-a, -a). A fifth particle that has a mass mand a charge +q is placed at the origin and released from rest. Find its speed when it is very far from the origin. (Use the following as necessary: k, q, m, and a.)
v =

Answer 1

Here we will use concept

Loss in electric potential energy=Gain in kinetic energy.

Loss in potential energy=(kq)/(2sqrt(a)) +(3kq)/(2sqrt(a)) +(-4kq)/(2sqrt(a)) +(10kq)/(2sqrt(a)) +(kq)/(2sqrt(a))

Loss in potential energy=(10kq)/(2sqrt(a))

So loss in potential energy=gain in kinetic energy=(0.5)(mv²)

(0.5)(mv²)=(10kq)/(2(sqrt(a))

v=sqrt((10kq)/(m×sqrt(a))

sqrt represents square root.