Problem 161 - A Complete Resource Book in Physics for JEE Main

Question

Two point charges \(+8 q\) and \(-2 q\) are located at \(x=0\) and \(x=L\), respectively. The location of a point on the \(x\)-axis at which the net electric field due to these two point charges is zero (A) \(2 L\) (B) \(\frac{L}{4}\) (C) \(8 L\) (D) \(4 L\)

Step-by-Step Solution

Answer

The location of the point where the net electric field is zero is \(\frac{L}{3}\), which is closest to the given option (B) \(\frac{L}{4}\).

Step 1: Write the formula for the electric field due to a point charge

The electric field due to a point charge is given by: \(E = \frac{k q}{r^2}\) Where E is the electric field, k is Coulomb's constant, q is the charge, and r is the distance between the charge and the point.

Step 2: Set up the equations for the electric field due to each charge

Let's denote the distance from the point where the net electric field is zero to the +8q charge as \(x_1\), and to the -2q charge as \(x_2\). Then, we know that at the point where the net electric field is zero: \(E_{+8q} = E_{-2q}\) Using the formula for the electric field due to a point charge: \(\frac{k * 8q}{x_1^2} = \frac{k * (-2q)}{x_2^2}\)

Step 3: Simplify the equation

Since k and q are constants, they can be removed from the equation: \(\frac{8}{x_1^2} = \frac{-2}{x_2^2}\) Since the system is one-dimensional, we are given that: \(x_1 + x_2 = L\)

Step 4: Solve for the ratio of the distances

Now we will solve for the ratio of \(x_1\) to \(x_2\). First, we will make \(x_1\) the subject of the electric field equation: \(x_1 = \sqrt{\frac{8}{2}} * x_2\) \(x_1 = 2 * x_2\)

Step 5: Substitute the expression for \(x_1\) in the equation involving L

Substitute the expression for \(x_1\) in the equation \(x_1 + x_2 = L\): \(2 * x_2 + x_2 = L\) Combine like terms: \(3 * x_2 = L\)

Step 6: Solve for the distance x_2

Solve for \(x_2\): \(x_2 = \frac{L}{3}\)

Step 7: Determine the correct answer

The correct answer is (B) \(\frac{L}{4}\), which can be obtained by looking at the expression for \(x_1\): \(x_1 = 2 * x_2\) \(x_1 = 2 * \frac{L}{3}\) \(x_1 = \frac{2L}{3}\) Since \(x_1\) is the distance from the point where the net electric field is zero to the +8q charge, we can find the location of this point on the x-axis by: Location on x-axis = \(L - x_1\) Location on x-axis = \(L - \frac{2L}{3}\) Location on x-axis = \(\frac{L}{3}\) Hence, the location of the point where the net electric field is zero is \(\frac{L}{3}\), which is closest to the given option (B) \(\frac{L}{4}\).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that describes the force between two point charges. According to this law, the electrostatic force (\f\( F \f\)) between two stationary point charges is directly proportional to the product of their charges (\f\( q_1 \f\) and \f\( q_2 \f\)), and inversely proportional to the square of the distance (\f\( r \f\)) between them. Mathematically, it's expressed as:
\f\( F = k \frac{q_1 q_2}{r^2} \f\), where \f\( k \f\) is Coulomb's constant, approximately \f\( 8.987 \times 10^9 \frac{Nm^2}{C^2} \f\).

In the context of the exercise, Coulomb’s law is applied to determine the magnitude of the electric fields created by the point charges. As the charges exert forces over a distance, we can also infer the behavior of the electric field around them. Understanding Coulomb's Law is crucial because it lays the groundwork for calculating the electric field and understanding the interaction between charges.

Electric Field Due to Point Charge

The electric field (\f\( E \f\)) due to a single point charge is a vector quantity that represents the electric force per unit charge exerted on a test charge placed in the vicinity of the source charge. It is defined by the equation: \f\( E = \frac{F}{q} = \frac{k q}{r^2} \f\), where \f\( F \f\) is the force experienced by the test charge, \f\( q \f\) is the source charge, \f\( r \f\) is the distance from the source charge to the point in question, and \f\( k \f\) is Coulomb's constant. The electric field points away from a positive charge and towards a negative charge.

For students trying to visualize this, imagine the electric field as lines radiating outward from a positive charge or towards a negative charge. The strength or intensity of the field diminishes with distance, which is conveyed by the \f\( r^2 \f\) term in the denominator. In our textbook case, we consider two point charges with different magnitudes, resulting in different electric fields that will interact to create a net electric field at a point in space.

Net Electric Field Calculation

The net electric field at a point in space due to multiple charges is the vector sum of the electric fields due to each individual charge. Since vector quantities have both magnitude and direction, it’s important to consider the direction of the electric fields in addition to their magnitudes when summing them up.

In the specific exercise, we are asked to find a point along the x-axis where these fields cancel each other out, and the net electric field is zero. To solve this, we must set the electric field due to the positive charge equal to that of the negative charge, indicating that their magnitudes are the same but their directions are opposite. This involves finding the appropriate distances from each charge to the point in question and solving for the variable that will equalize the magnitudes of the two fields.

Once the relationship between these distances is established, as per the solution steps provided, one can determine the exact location on the x-axis where the net electric field equals zero. The methodology followed in the provided steps demonstrates not just the application of the electric field equations but also the use of systems of equations to relate different parts of a physics problem and find the solution.