Three charges, each +Q, are placed at the three corners of a square of side L. The magnitude of electric field at the fourth corner of the square will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (2√2 + 1) Q/(8 π ε0 L2)
The field at the empty corner is the vector sum (superposition) of the fields of the three charges. Let \( k = 1/4\pi\varepsilon_0 \).
The two charges on the adjacent corners are at distance L; their fields are equal and at right angles:
\( \displaystyle E_1 = E_2 = \frac{kQ}{L^{2}} \)
\( \displaystyle E_{12} = \sqrt{E_1^{2} + E_2^{2}} = \frac{\sqrt{2}\,kQ}{L^{2}} \)
This resultant points along the diagonal, away from the square. The charge at the opposite corner is at distance \( \sqrt{2}L \), and its field acts along the same diagonal in the same direction:
\( \displaystyle E_3 = \frac{kQ}{(\sqrt{2}L)^{2}} = \frac{kQ}{2L^{2}} \)
Adding the two collinear fields:
\( \displaystyle E = \frac{kQ}{L^{2}}\left(\sqrt{2} + \frac{1}{2}\right) = \frac{(2\sqrt{2} + 1)\,kQ}{2L^{2}} \)
\( \displaystyle E = \frac{(2\sqrt{2} + 1)\,Q}{8\pi\varepsilon_0 L^{2}} \)