Question

Equal charges Q are placed at the four corners A, B, C, D of a square of length a. The magnitude of the force on the charge at B will be

• A. $\frac{3Q^{2}}{4\pi\varepsilon_{0}a^{2}}$
• B. $\frac{4Q^{2}}{4\pi\varepsilon_{0}a^{2}}$
• C. $\left( \frac{1 + 1\sqrt{2}}{2} \right)\ \frac{Q^{2}}{4\pi\varepsilon_{0}a^{2}}$
• D. $\left( 2 + \frac{1}{\sqrt{2}} \right)\ \frac{Q^{2}}{4\pi\varepsilon_{0}a^{2}}$

Answer 1

Answer: (C)

$\left( \frac{1 + 1\sqrt{2}}{2} \right)\ \frac{Q^{2}}{4\pi\varepsilon_{0}a^{2}}$

Answer 2

After following the guidelines mentioned above

$F_{net} = F_{AC} + F_{D} = \sqrt{F_{A}^{2} + F_{C}^{2} +}F_{D}$Since $F_{A} = F_{C} = \frac{kQ^{2}}{a^{2}}$and $F_{D} = \frac{kQ^{2}}{(a\sqrt{2})^{2}}$

$F_{net} = \frac{\sqrt{2}kQ^{2}}{a^{2}} + \frac{kQ^{2}}{2a^{2}} = \frac{kQ^{2}}{a^{2}}\left( \sqrt{2} + \frac{1}{2} \right) = \frac{Q^{2}}{4\pi\varepsilon_{0}a^{2}}\left( \frac{1 + 2\sqrt{2}}{2} \right)$