Question

Four electric charges \(+q, +q, -q\) and \(-q\) are placed at the corners of a square of side \(2L\) (see figure). The electric potential at point \(A\), midway between the two charges \(+q\) and \(+q\), is

• A. $ \frac{1}{ 4 \pi \varepsilon_0 } \frac{ 2q}{ L} (1 + \sqrt 5)$
• B. $\frac{1}{ 4 \pi \varepsilon_0 } \frac{ 2q}{ L} \bigg [ 1 - \frac{ 1}{ \sqrt 5} \bigg ] $
• C. $ \frac{1}{ 4 \pi \varepsilon_0 } \bigg(1 + \frac{ 2q}{ L} (1 - \frac{1}{\sqrt 5})\bigg)$
• D. zero

Answer 1

The Correct Option is C

To find the electric potential at point \( A \), we need to consider the contributions from each of the four charges located at the corners of the square. The side of the square is \( 2L \), and point \( A \) is midway between the two \( +q \) charges, which means point \( A \) is at a distance \( L \) from each of those charges.

1. The electric potential due to a single point charge \( q \) at a distance \( r \) is given by: V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}.
2. At point \( A \), the potential due to each \( +q \) charge is: V_{+q} = \frac{1}{ 4\pi \varepsilon_0 } \frac{q}{L}.
3. The distance from point \( A \) to each \( -q \) charge at the opposite corners is: \sqrt{(2L)^2 + (L)^2} = L\sqrt{5}.
4. The potential due to each \( -q \) charge is: V_{-q} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{L\sqrt{5}}.
5. Total potential at \( A \) is the algebraic sum of potentials due to all charges:
   V_A = 2V_{+q} + 2V_{-q}
   = 2 \left(\frac{1}{4\pi \varepsilon_0} \frac{q}{L}\right) + 2 \left(\frac{1}{4\pi \varepsilon_0} \frac{-q}{L\sqrt{5}}\right)
   = \frac{1}{4\pi \varepsilon_0} \left( \frac{2q}{L} - \frac{2q}{L\sqrt{5}} \right)
   = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2q}{L} \left( 1 - \frac{1}{\sqrt{5}} \right).

Thus, the correct answer is:

\(\frac{1}{ 4 \pi \varepsilon_0 } \bigg(1 + \frac{ 2q}{ L} (1 - \frac{1}{\sqrt 5})\bigg)\)