Problem 46 - A Complete Resource Book in Physics for JEE Main

Question

The magnitude of electric intensity at a distance \(x\) from a charge \(q\) is \(E\). An identical charge is placed at a distance \(2 x\) from it. Then the magnitude of the force it experiences is (A) \(q E\) (B) \(2 q E\) (C) \(\frac{q E}{2}\) (D) \(\frac{q E}{4}\)

Step-by-Step Solution

Answer

The magnitude of the force experienced by the second charge is \(\frac{q E}{4}\), which corresponds to option (D).

Step 1: Write the formula for electric field intensity

The electric field intensity (E) at a distance (x) from a charge (q) can be given by the formula: \(E = \frac{kq}{x^2}\), where \(k\) is Coulomb's constant, approximately \(8.99 × 10^9 N·m^2/C^2\).

Step 2: Calculate the force between the two charges

To find the force between two charges separated by distance, we use Coulomb's law: \(F = \frac{kq1q2}{r^2}\). In our case, both charges are equal (\(q1 = q2 = q\)), and the separation distance is \(2x\). Therefore, the formula becomes: \(F = \frac{kq^2}{(2x)^2}\)

Step 3: Substitute E in the equation for force

We will now substitute the electric intensity (E) into the equation for F, using the equation from Step 1: \(E = \frac{kq}{x^2}\) First, we solve \(kq\) in terms of E and x: \(E x^2 = kq \Rightarrow kq = E x^2\) Now, we replace \(kq\) in the equation for F: \(F = \frac{(Ex^2)q}{(2x)^2}\)

Step 4: Simplify the equation and find the force

The equation for the force can be simplified as follows: \(F = \frac{(E x^2)q}{(4x^2)}\) Notice that \(x^2\) can be canceled from the numerator and the denominator: \(F = \frac{Eq}{4}\) Hence, the magnitude of force experienced by the second charge is: \(F = \frac{q E}{4}\) Therefore, the correct choice is (D) \(\frac{q E}{4}\).