Question

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?
2) Where would the net electric field be zero if one of the charges were negative?

Answer 1

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C   and q₂ = 11.0 n C

assume the distance be r from q₁  where the electric field is zero.

distance of point from q₂  be equal to 1.8 -r

now,

        E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

b) zero when one charge is negative.

let us assume  q₁  be negative so, distance from  q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

   E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m