Question

Two particles are fixed to an x axis: particle 1 of charge -3.91 x 10^-7 c is at the origin and particle 2 of charge +3.91 x 10^-7 c is at x2 = 12.8 cm. Midway between the particles, what the magnitude of the net electric field?

Answer 1

Answer:

Net electric field, $E_{net}=4.29\times 10^5\ N/C$

Explanation:

It is given that,

Charge on particle 1, $q_1=-3.91\times 10^{-7}\ C$

Charge on particle 2, $q_2=+3.91\times 10^{-7}\ C$    

Particle 1 is at origin and particle 2 is at x₂ = 12.8 cm = 0.128 m

The magnitude of the net electric field at the midway between the particles is given by :

$E=\dfrac{kQ}{x^2}$

$E=\dfrac{9\times 10^9\times 3.91\times 10^{-7}}{(0.128\ m)^2}$

E = 214782.71 N/C

We know that for a positive charge the direction of field is outwards and for a negative charge the direction of electric field is inwards. Here, for two charges, the net electric field,

$E_{net}=2\times E$

$E_{net}=2\times 214782.7 N/C$

$E_{net}=429565.42\ N/C$

$E_{net}=4.29\times 10^5\ N/C$

So, the magnitude of the net electric field is $4.29\times 10^5\ N/C$. Hence, this is the required solution.