Consider two charges q = 4.15 nC at x = 0 cm and q2 = 2.61 nC at x2 = 30.5 cm. A point A is a distance d = 12.9 cm to the right of q2. Point B is the same distance d = 12.9 cm to the right of 91. a) Find the electric field vector due to q at A: = b) Find the electric field vector E2 due to q at A: = c) Find the electric field vector E due to both charges at A: A = d) Find the electric field vector E1B due to q1 at B: E1B = e) Find the electric field vector E2B due to q2 at B: E2B = F) Find the electric field vector EB due to both charges at B: EB = g) At what distance to the right of 91 would the electric field due to both charges be equal to zero? x = We replace q by an equal negative charge 91 = -4.15 nC h) Find the new electric field vector due to both charges at A: A = A i) Find the new electric field vector EB due to both charges at B: B =