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 is:

• A. \( \left(\frac{4d}{3}, 0, 0\right) \)
• B. \( \left(\frac{d}{4}, 0, 0\right) \)
• C. \( \left(\frac{3d}{4}, 0, 0\right) \)
• D. \( \left(\frac{d}{3}, 0, 0\right) \)

Hint:

In problems involving multiple charges, use the principle of superposition for electric fields and set the total field equal to zero to find the point of cancellation.

Answer 1

The Correct Option is B

Let the electric field at point \( P \) in between the charges be zero. Let the position of \( P \) be at a distance \( x \) from the origin, where the electric field due to both charges cancels each other.  
The electric field due to a point charge is given by: \[ E = \frac{kq}{r^2} \] For the electric field to be zero at point \( P \), the fields due to both charges must be equal and opposite. So: \[ \frac{kq}{x^2} = \frac{k(9q)}{(d - x)^2} \] Simplifying: \[ \frac{1}{x^2} = \frac{9}{(d - x)^2} \] Solving for \( x \): \[ d - x = 3x \quad \Rightarrow \quad d = 4x \quad \Rightarrow \quad x = \frac{d}{4} \] Thus, the coordinate of point \( P \) is \( \left(\frac{d}{4}, 0, 0\right) \).

Answer 2

The problem asks to find the point on the line segment between two point charges, \(+q\) at the origin and \(+9q\) at \((d, 0, 0)\), where the net electric field is zero.

Concept Used:

This problem is solved using the principle of superposition for electric fields. The electric field at a point due to a system of charges is the vector sum of the electric fields produced by each charge individually.

The magnitude of the electric field \( E \) at a distance \( r \) from a point charge \( Q \) is given by Coulomb's law:

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

where \( k \) is Coulomb's constant. For the net electric field to be zero at a point, the electric fields from all charges at that point must sum to zero. Since both charges are positive, the point where the field vanishes must lie on the line segment connecting them. At this point, the electric field from the first charge will be equal in magnitude and opposite in direction to the electric field from the second charge.

Step-by-Step Solution:

Step 1: Set up the coordinate system and define the point of interest.

Let the first charge be \( q_1 = +q \) located at the origin \( (0, 0, 0) \). Let the second charge be \( q_2 = +9q \) located at \( (d, 0, 0) \).

Let the point where the electric field vanishes be \( P \), with coordinates \( (x, 0, 0) \). Since both charges are positive, the fields can only cancel out at a point between them. Therefore, we must have \( 0 < x < d \).

Step 2: Write down the expressions for the electric fields from each charge at point P.

The electric field \( \vec{E}_1 \) at point P due to charge \( q_1 \) is directed along the positive x-axis. Its magnitude is:

\[ E_1 = k \frac{q}{x^2} \]

The electric field \( \vec{E}_2 \) at point P due to charge \( q_2 \) is directed along the negative x-axis. Its magnitude is:

\[ E_2 = k \frac{9q}{(d - x)^2} \]

where the distance from \( q_2 \) to P is \( d - x \).

Step 3: Equate the magnitudes of the electric fields.

For the net electric field at point P to be zero, the magnitudes of \( \vec{E}_1 \) and \( \vec{E}_2 \) must be equal:

\[ E_1 = E_2 \] \[ k \frac{q}{x^2} = k \frac{9q}{(d - x)^2} \]

Final Computation & Result:

Step 4: Solve the equation for \( x \).

We can cancel \( kq \) from both sides of the equation:

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

Taking the square root of both sides:

\[ \frac{1}{x} = \pm \frac{3}{d - x} \]

Since we know the point \( P \) lies between the two charges (\( 0 < x < d \)), both \( x \) and \( d - x \) are positive. Thus, we must take the positive root:

\[ \frac{1}{x} = \frac{3}{d - x} \]

Now, we cross-multiply to solve for \( x \):

\[ d - x = 3x \] \[ d = 4x \] \[ x = \frac{d}{4} \]

This value of \( x \) satisfies the condition \( 0 < x < d \).

Therefore, the point where the electric field vanishes is \( \left(\frac{d}{4}, 0, 0\right) \).