Question

Find the net electric field at point A​

Answer 1

Answer:

Enet = √3q x 10¹³ N/C

Explanation:

Eb = Kq/(3x 10^-2)² = q x 10¹³ = E

Ec = K x 2q/(3x 10^-2)² = 2q x 10¹³ = 2E

angle between Eb and Ec is 60 degree

now, ___ __ __

Enet = Eb + Ec

magnitude--

Enet = √{ (Eb square)+(Ec square) - 2EbEc cos60}

Enet = √{ E² + (2E)² - 2(E)(2E)x 1/2}

Enet = √3q x 10¹³ N/C