Question

two particles having charges q1=0.500nC and q2 8.00nC are separated by a distance of 1.20m. at what point along the line connecting the two charges is the total electric field due to the two charges equal zero?

Answer 1

Given : q₁ = 0.500 ×10^-9 C , q ₂ = 8 × 10^-9 C , d = 1.20 m

Goal : The point along the line connecting the two charges ,at which the electric field is zero.

Solution :

Electric field at a point due to a point charge q is given by , E = kq/r² , where r is the distance.

We know that , the electric field at a point would be zero if electric field due to charge q₁ is equal to the electric field due to charge q₂ at that point.

Suppose the point at which electric field is zero , is at a distance x from q₁

Then electric field at that point due to q₁ is :

E₁ = kq₁/x²

And Electric field at that point due to q₂ is :

E₂= kq₂/(1.20-x)²

E₁ = E₂

kq₁/x² = kq₂/(1.20-x)²

(1.20-x)² = (q₂/q₁)x²

x² - 2.4x + 1.44 = (8/0.5)x²

15x² + 2.4x - 1.44 = 0

By solving quadratic equation , we get x = 0.24 , -0.4

As x can not be negative , so x = 0.24.

Hence the total electric field due to two charges is zero at 0.24 m from q₁ or 0.96 m (1.20- 0.24) from q₂