Two point charges are inserted on the x axis. (Figure 1)The firstcharge, q1 = 8.00 nC , is inserted a street 16.0 m native the originalong the hopeful x axis; the 2nd charge, q2 = 6.00 nC , isplaced a street 9.00 m native the origin along the an adverse xaxis.

An unknown extr charge q3 is now put at point B,located at collaborates (0 m, 15.0 m ). Find the magnitude and also signof q3 essential to do the full electric ar at suggest A equal tozero.

You are watching: Calculate the electric field at point a, located at coordinates (0 m, 12.0 m ).

Position vector native ql, F-16 m,12 m>(-16m)'+(12m-20m place vector from q2, = now, let electric field at suggest A,

E+E2+E-0 r2 Now, lets plug every values, 20m) (15m) 0.18「+0.24ん+10"q3r3:0 0.18+0.24 +109q,=0 ++=0 net electric field along y axis,

Position vector native ql, F-16 m,12 m>(-16m)'+(12m-20m place vector from q2, = now, let electric field at suggest A,

E+E2+E-0 r2 Now, lets plug every values, 20m) (15m) 0.18「+0.24ん+10"q3r3:0 0.18+0.24 +109q,=0 ++=0 net electric field along y axis,

Cos theta_1 = 16/20 = 4/5 cos theta_2 = 3/5 q_1 = 8nc q_2 = 6nc E_1 = 1/4 pi epsilon_0 q_1/(20)^2 E_1 cos theta_1 = 1/4 pi epsilon_0 q_1/(20)^2 4/5 = 1/4 pi epsilon_0 8 times 4/(20)^2 times 5 = 0.016/4 pi epsilon_0 E_2 = 1/4 pi epsilon_0 q_2/(15)^2 E_2 cos theta_2 = 1/4 pi epsilon_0 q_2/(15)^2 3/5 = 1/4 pi epsilon_0 6/(15)^2 times 3/5 = 0.016/4 pi epsilon_0 at suggest A E^vector because of x contents is same and also cancelled so over there is just component in direction that y ar so E_net^vector = E_q_3^vector E_1 sin theta_1 + E_2 sin theta_2 = 1/4 pi epsilon_0 q_3/(3)^2 1/4 pi epsilon_0 (q_1/(20)^2) time 3/5 + 1/4 pi epsilon_0 q_2/(15)^2 times 4/5 = 1/4 pi epsilon_0 q_3/(3)^2 \r\n q_1/400 time 3/5 + q_2 time 4/225 times 5 = q_3/9 0.012 + 0.0214 = q_3/9 q_3 = 0.3nc

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R_1 = 15 m, r_2 = 20 m r = squareroot 15^2 + 20^2 = 25 m sin theta = 12/20 = theta_r_1 = 36.86 degreesin theta_12 = 12/15 theta_q_2 = 53.13 level f_1 = k r_1/r_1^2 = 9 timees 10^3 times 8 time 10^-3/20^2 = 0.18 n/c f_2 = k q_2/r^2_2 = 9 time 10^9 time 6 times 10^-5/15^2 = 0.24 N/c f_1 x = 0.18 times cos (-36.87) = -0.143 N/c f_2 y = 0.18 sin (36.87) = 0.108 N/c f_2 x = 0.24 cos (53.13) = 0.144 N/c f_2 y = 0.24 sin (53.13) = 0.192 N/c f_x = -0.143 + 0.144 = 0 N/c f_y = 0.108 + 0.152 = 0.3 N/c EA = (10, 0.3) -a (d) FA = squareroot (f_1 x + f_2 x)^2 + (f_1 y + f_2 y)^2 = squareroot 0.09 = 0.3 N/c r = 15 -12 = 3 m FA = k r b/r^2 0.3 =