Question

A +8.0 nC charge is located at (x,y)=(0cm,13cm) and a -7.0 nC charge is located at (x,y)=(5.0cm,0cm).

Where would a -15 nC charge need to be located in order that the electric field at the origin be zero?

Answer 1

Answer :

Calculate the E field at 0,0 and then place the -15nC charge to exactly oppose it in magnitude and direction by adjusting it's distance to be equal in magnitude and it's location to be exactly 180 different in direction. Lets do it:
The directions of the fields at the origin due to the 8nC and -7nC charges are:
The +8nC field points at 270° . The -7nC field at 0° because + charge fields point OUT and - charge fields point IN
E=kq/r²
Plug in values: .
Ex = k*7e-9/0.0025 = 11,005.7 N/C at 0°
Ey = k*8e-9/0.0169 = 4993.3 N/C at 270°
The fields are vectors and perpendicular so use Pythagoras theorem:
E = √Ex² + Ey² ) = 14315.7 at 335.59° = arctan(Ey/Ex)
So the location of the -15nC will be 335.59°-180° = 155.59° because the E field from the -15nC charge points INTO the -15nC charge in exactly the opposite direction to the field at the origin
The distance from the origin will be r² = k*15e-9/14315.7 => r = 9.710cm
y = 9.710*sin155.59 = 4.013 cm
x = 9.710*cos155.59 = -8.84cm  

(x,y) = (-8.84, 4.013) cm