Question

The figure shows two protons separated by a distant d. Point Pis a distance d/4 from the leftmost charge. If d is equal to 2 nm, what is the magnitude of the net electric field at point P, and which way does it point? proton d/4 P proton 3.3E9 N/C, to the right O 7.2E9N/C, to the left 7.2E9 N/C, to the right 5.1E9 N/C, to the left O 5.1E9 N/C, to the right

Answer 1

Answer :  ◉ 5.1E9 N/C, to the right.

Solution :

Here we have :

d = 2 nm = 2 x 10^-9 m

And, Charge of proton is : q = 1.6 x 10^-19 C

.

1 2 E2 E1 q a d/4- 3d/4

.

Here, Net electric field at point P will be given by :

E_net = E₁ - E₂

_{$\therefore \ E_{net}=\frac{kq}{(d/4)^2}-\frac{kq}{(3d/4)^2}$}

_{$\therefore \ E_{net}=\frac{16kq}{(d)^2}-\frac{16kq}{(9)(d)^2}$}

_{$\therefore \ E_{net}=\frac{16kq}{(d)^2}\left [ 1-\frac{1}{9} \right ]$}

_{$\therefore \ E_{net}=\frac{16kq}{(d)^2}\left [\frac{8}{9} \right ]$}

_{$\therefore \ E_{net}=\frac{128}{9}\cdot \frac{kq}{(d)^2}$}

_{$\therefore \ E_{net}=\frac{128}{9}\cdot \frac{(9\times 10^9)(1.6\times 10^{-19})}{(2\times 10^{-9})^2}$}

_{$\therefore \ E_{net}=\mathbf{5.12\times 10^9\ N/C}$}

And, The direction of the electric field will be to the right.