For a given dipole at a poit (away from the center of dipole) intensity of the electric field is E. Charges of the dipole are brought closer such that distance between point charges is half, and magnitude of charges are also halved. The intensity of the field now at the same point becomes

Options

A. One fourth
B. Doubled
C. Four times
D. Halved

Answer

To solve the problem step by step, we will analyze the situation before and after the changes made to the dipole.

### Step 1: Understand the Initial Conditions
Initially, we have a dipole consisting of two charges, \( +Q \) and \( -Q \), separated by a distance \( d \). The dipole moment \( P \) is given by:
\[ P = Q \cdot d \]
The electric field \( E \) at a point away from the dipole can be expressed as:
\[ E = \frac{kP}{D^3} \]
where \( k \) is Coulomb's constant, \( D \) is the distance from the center of the dipole to the point where the electric field is being measured, and we assume \( D \gg d \).

### Step 2: Analyze the Changes Made to the Dipole
The problem states that the distance between the charges is halved, and the magnitude of the charges is also halved. Therefore, the new distance between the charges becomes:
\[ d' = \frac{d}{2} \]
The new charges are:
\[ Q' = \frac{Q}{2} \]
Now, we can calculate the new dipole moment \( P' \):
\[ P' = Q' \cdot d' = \left(\frac{Q}{2}\right) \cdot \left(\frac{d}{2}\right) = \frac{Qd}{4} = \frac{P}{4} \]

### Step 3: Calculate the New Electric Field
The electric field \( E' \) at the same point after the changes can now be expressed as:
\[ E' = \frac{kP'}{D^3} = \frac{k \left(\frac{P}{4}\right)}{D^3} = \frac{kP}{4D^3} \]

### Step 4: Relate the New Electric Field to the Original Electric Field
We know that the original electric field \( E \) was:
\[ E = \frac{kP}{D^3} \]
Now, substituting this into the equation for \( E' \):
\[ E' = \frac{1}{4} \cdot \frac{kP}{D^3} = \frac{E}{4} \]

### Final Answer
Thus, the intensity of the electric field at the same point after the changes is:
\[ E' = \frac{E}{4} \]