Question

Point charges of -6 mue C, and +6 mue C are located along the x axis x=-1.0 cm, x=+1.0 cm, respectively. Calculate the electric field at x=3.0 cm. Locate a point on the positive x axis where the magnitude of field is zero.

Answer 1

E.F. at x=3 cm due to -6$\mu$C , E1= -6x10^-6x9x10⁹/ (4x10^-2)² N/C

E.F. at x=3 cm due to 5$\mu$C , E2= 5x10^-6x9x10⁹/ (3x10^-2)² N/C

E.F. at x=3 cm due to 6$\mu$C , E3= 6x10^-6x9x10⁹/ (2x10^-2)² N/C

Total E.F. at x=3 cm, E = E1+E2+E3 = 15.12 x 10⁷ N/C

E.F. at x=15 cm due to -6$\mu$C , E4= -6x10^-6x9x10⁹/ (16x10^-2)² N/C

E.F. at x=15 cm due to 5$\mu$C , E5= 5x10^-6x9x10⁹/ (15x10^-2)² N/C

E.F. at x=15 cm due to 6$\mu$C , E6= 6x10^-6x9x10⁹/ (14x10^-2)² N/C

Total E.F. at x=15 cm, E = E4+E5+E6 = 0.27 x 10⁷ N/C

Say at x cm the E.F. is zero.

so we can write

9x10⁹x 10^-6 x10⁴[-6/(x+1)²+5/x²+6/(x-1)²]=0

Solving this we get x=0.69 cm