82 points CJ9 18P045 Ask Your Teacher My Notes Question Part

8.–/2 points CJ9 18.P.045. Ask Your Teacher My Notes Question Part Points Submissions Used Two point charges are located along the x axis: q1 = +6.2 C at x1 = +3.8 cm, and q2 = +6.2 C at x2 = 3.8 cm. Two other charges are located on the y axis: q3 = +3.1 C at y3 = +4.9 cm, and q4 = 7.9 C at y4 = +6.9 cm.

Find the net electric field (magnitude and direction) at the origin.

Solution

field due to a point charge at distance r,

E = k q / r^2

away from+ve charge and towards the -ve charge.

due to q1:

E1 = (9 x 10^9 x 6.2 x 10^-6) / (3.8 x 10^-2)^2 (-i)

E1 = - 3.86 x 10^7 N/c i

due to q2:

E2 = (9 x 10^9 x 6.2 x 10^-6) / (3.8 x 10^-2)^2 (-i)

E2 = - 3.86 x 10^7 N/c i

due to q3:

E3 = (9x 10^9 x 3.1 x 10^-6 ) / (4.9x 10^-2) (-j)

E3 = - 1.16 x 10^7 N/C j

due to q4:

E4 = (9x 10^9 x 7.9 x 10^-6 ) / (6.9x 10^-2) (-j)

E4 = - 1.49 x 10^7 N/C j   

E = E1 + E2 + E3 + E4

E = - 7.72 x 10^7 N/C i - 2.65 x 10^7 N/C j

magnitude = sqrt(7.72^2 + 2.65^2) x 10^7 = 8.16 x 10^7 N/C .....Ans

Direction:

@ = 180 + tan^-1(2.65 / 7.72) = 199 deg counter clock wise to x axis.