Two charges are arranged at corners of a square which has a side length of L = 0.25 m. The values of the charges are q1 = −3.3 × 10−6 C and q2 = +4.2 × 10−6 C. (a) Find the electric potential at points A and B. (b) If a third charge q3 = +2.1 × 10−6 C is placed at corner A, determine the electric potential energy of the system.

a) Electric potential at Point A=66342.14v

Point B= -10800v

b) 928800J

Explanation:

Diagonal of the square= 0.25×1.414=0.35m^2

CornerA=kq1/A= (9×10^9)×(-3.3×10^-6)/0.35

=29700/0.35=-84857.14V

Kq2/A= (9×10^9)(4.2×10^-6)/0.25=37800/0.25=151200V

Sum= (-84857.14+151200)V= 66342.86v

Point B

Kq1/0.25= (9×10^9)(-3.3×10^-6)/0.25= -118800v

Kq2/0.35= (9×10^9)(4.2×10^-6)/0.35= 108000v

Sum= -118800+ 108000= -10800V

b) PE = q3×(sum of v)=2.1×10^-6)× 66342.86

PE= 928800V

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

c. magnitude and direction