Question

Point charge of 10 µC is placed at the origin. At what location on the X-axis should a point charge of 40 µC be placed so that the net electric field is zero at x = 2 cm on the X-axis?

Options

• x = 6 cm

• x = 4 cm

• x = 8 cm

• x = −4 cm

Answer 1

x = 6 cm

Explanation:

Given: Charge (q₁) = +10 μC at origin x = 0

Charge (q₂) = +40 μC at x = ?

Net electric field = 0 at x = 2 cm

At the point where the net field is zero:

E₁​ = E_2​

`(k q_1)/r_1^2 = (k q_2)/r_2^2`    ...(i)

Distance from q₁:

r₁​ = 2 cm

Let q₂ be at position x, so the distance:

r₂​ = |x − 2|

Substituting all the values in equation (i), we get,

`10/(2)^2 = 40/((x - 2)^2)`

`10/ 4 = 40/((x - 2)^2)`

`5/2 = 40/((x - 2))`

(x − 2)² = `(40 xx 2)/5`

= `80/5`

(x − 2)² = 16    ...[Taking the square root on both sides]

(x − 2) = ±4

x = 4 + 2 or x = −4 + 2

x = 6 or x = −2

At x = 2 cm, both charges are positive, so the fields must be in opposite directions.

At x = 6, the fields oppose it is valid.

At x = −2, the fields are in the same direction, which is not valid.