Question

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes

• A. (d/4, 0, 0)

Answer 1

Answer: (A)

(d/4, 0, 0)

Answer 2

To find the point where the electric field vanishes between two positive charges (+q and +9q), we need to determine the location where the electric fields due to each charge cancel each other out.

Let the point be at a distance $x$ from the origin (where +q is located) along the x-axis. Therefore, the distance from the second charge +9q (located at $d$) will be $d - x$.

The electric field $E$ due to a point charge $Q$ at a distance $r$ is given by:

$E = k \frac{|Q|}{r^2}$

where $k$ is the electrostatic constant.

1. Electric field due to +q ($E_1$): $E_1 = k \frac{q}{x^2}$

2. Electric field due to +9q ($E_2$): $E_2 = k \frac{9q}{(d-x)^2}$

For the net electric field to be zero:

$E_1 = E_2$

$k \frac{q}{x^2} = k \frac{9q}{(d-x)^2}$

We can cancel out $k$ and $q$:

$\frac{1}{x^2} = \frac{9}{(d-x)^2}$

Taking the square root of both sides:

$\frac{1}{x} = \frac{3}{d-x}$

Solving for $x$:

$d - x = 3x$

$d = 4x$

$x = \frac{d}{4}$

Thus, the point where the electric field vanishes is at $(\frac{d}{4}, 0, 0)$.