A 6.00 nC is 2.00 m from a 3.00 nC charge. Find the magnitude of the electric field at a point midway between the two charges? Which way does the electric field point, towards the positive or the negative charge?

E_total = 26.97 N/C

Electric field points towards the positive charge

Explanation:

We are given;

Charge 1; q1 = 6 nC = 6 × 10^(-9) C

Charge 2; q2 = 3 nC = 3 × 10^(-9) C

Distance between both charges; R_o = 2 m

Since we want to find electric field midway, the distance midway is r = 2/2 = 1 m

Using coulumbs law;

E = kq/r²

Where k is a constant with a value of 8.99 × 10^(9) N.m/C²

Thus;

E1 = kq1/r²

E1 = (8.99 × 10^(9) × 6 × 10^(-9))/1²

E1 = 53.94 N/C

Similarly;

E2 = kq2/r²

E2 = (8.99 × 10^(9) × 3 × 10^(-9))/1²

E2 = 26.97 N/C

Since both electric fields are positive, it means that they are both moving towards the midpoint of the distance between both charges.

This implies they will have opposite directions.

Thus, total electric field at the midway point is;

E_total = E1 - E2

E_total = 53.94 - 26.97

E_total = 26.97 N/C

real, smaller and upside-down.   so the answer should be b.

answera compound;