Question

Three charges, each $+q$, are placed a at the corners of an isosceles triangle $ABC$ of sides $BC$ and $AC$, $2a$. $D$ and $E$ are the mid points of $BC$ and $CA$. The work done in taking a charge $Q$ from $D$ to $E$ is

• A. $ \frac {3 q Q}{ 4 \pi \varepsilon_0 a } $
• B. $ \frac{ 3 q Q}{ 8 \pi \varepsilon_0 a } $
• C. $ \frac{ q Q}{ 4 \pi \varepsilon_0 a } $
• D. zero

Answer 1

The Correct Option is D

To determine the work done in moving a charge \( Q \) from point \( D \) to point \( E \), we need to understand the concept of electric potential. The work done in moving a charge in an electric field is given by the change in electric potential energy, which can be expressed as:

W = Q (V_E - V_D)

where \( V_E \) and \( V_D \) are the electric potentials at points \( E \) and \( D \), respectively.

Given that the charges are located at the vertices of an isosceles triangle, let's analyze the situation:

• Since \( D \) and \( E \) are midpoints of \( BC \) and \( CA \) respectively, they are equidistant from the charges at \( B \) and \( C \).
• The electric potential at any point due to a point charge \( +q \) is given by V = \frac{kq}{r}, where \( k \) is the Coulomb's constant.

Since the triangle is symmetrical, assume \( D \) and \( E \) lie along a line parallel to the base of the triangle and at the midpoint of sides.

The electric potential at \( D \) due to the charges at \( B \), \( C \), and \( A \) adds to zero in terms of difference because:

• Both \( D \) and \( E \) experience the same potential change from charges at \( A \) and the line of symmetry due to geometry.

Because both \( D \) and \( E \) are symmetrically located with respect to the charges, the potential difference \( V_E - V_D = 0 \).

Thus, the work done is:

W = Q \times 0 = 0

This means the work done in moving a charge \( Q \) from \( D \) to \( E \) is zero because there is no potential difference.

Therefore, the correct answer is zero.