Question

Four point charges have the same magnitude of 2.7 × 10-12 C and are fixed to the corners of a square that is 7.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Answer 1

We know that electric field is given by:

E = k|Q|/R^2

Where direction of electric field will be away from positive charge and towards the negative charge.

We are choosing coordinate system such that center of square is at the origin, and place charge in any order as per your convenience, then

Now Using above FBD, we can see that net electric field at the center of square will be:

E_net = E1 + E3

Since E2 and E4 are equal and in opposite direction, So net electric field in vertical direction will be zero

E_net = k*|q1|/r1^2 + k*|q3|/r3^2

|q1| = |q3| = 2.7*10^-12 C

r1 = r3 = distance between center of square and any corner = diagonal/2 = (a*sqrt 2)/2 = a/sqrt 2

r1 = r3 = 7.0 cm/sqrt 2 = (0.07/sqrt 2) m

So, Since r1 = r3, and |q1| = |q3|

E_net = 2*k*q1/r1^2

E_net = 2*9*10^9*2.7*10^-12/(0.07/sqrt 2)^2

E_net = 4*9*2.7*10^-3/0.07^2

E_net = 19.84 N/C = magnitude of net electric field

Let me know if you've any query.