A positive point charge of 3.1*10^(-7) C at the point (x, y) = (-1.5/2m, 0m) is

Question

A positive point charge of 3.1*10^(-7) C at the point (x, y) = (-1.5/2m, 0m) is 1.5m away from a negative point charge of 3.1*10^(-7) C at (x,y) = (+1.5/2m, 0m) on the x-axis. Find the electric field (size and directions) at a point on the y-axis (x,y) = (0m, 3.1m) due to the two charges.

a) 548.54 N/C, along the y-axis

b) 533.16 N/C, along the y-axis

c) 548.54 N/C, parallel to the x-axis

d) 0.00 N/C, since the field cancel

e) 128.99 N/C, parallel to the x-axis

Please give equations you used, thanks!

Explanation / Answer

First determine the direction of the E fields at the point 0,3.1 which I'll call P
From q1 = 3.1*10^(-7) the field at P is at arctan(0) =0°
Because the field points towards the negative charge.
From q2 = 3.1*10^(-7) charge the field at P is at arctan(0) = 0°+180° = 180°
Because the field points away from the positive charge
r1=r2=1.5m

E = kq/r²
E1 = (9x10^9)*3.1*10^(-7)/1.5*1.5= 12400 N/C
E2 = (9x10^9)*3.1*10^(-7)/1.5*1.5 = 12400 N/C

Sum the horizontal components
E1*cos0 + E2*cos180 = 24800
Sum the vertical components
E1*sin0 + E2*sin180 = 0 N/C
So the y components cancel and the x add. The direction of the E field is 270

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.