Question

Four charges + q each are located at the vertices of square ABCD of side a as shown in figure. Find the electric field E at the midpoint of side BC –

• A. $\frac{4}{5\sqrt{5}\pi \in_{0}}\frac{a^{2}}{q^{2}}$
• B. $\frac{4}{5\sqrt{5}\pi \in_{0}}\frac{q^{2}}{a^{2}}$
• C. 0
• D. None of the above

Answer 1

Answer: (B)

$\frac{4}{5\sqrt{5}\pi \in_{0}}\frac{q^{2}}{a^{2}}$

Answer 2

E = E₁ +E₂ + E₃ + E₄

The component of E₂ and E₃ along the line will cancel each other then

E = E₁ + E₄

The component perpendicular to BC will added up

AP = $\sqrt { \mathrm { AB } ^ { 2 } + \mathrm { BP } ^ { 2 } }$ == $\frac { a \sqrt { 5 } } { 2 }$

DP = $\frac { a \sqrt { 5 } } { 2 }$ cos q === $\frac { 2 } { \sqrt { 5 } }$

\ E_P = E₁ cosq + E₄ cosq

= 2 × $\frac { 1 } { 4 \pi \epsilon _ { 0 } }$ . $\frac { q } { \left( \frac { a \sqrt { 5 } } { 2 } \right) ^ { 2 } }$ . $\frac { 2 } { \sqrt { 5 } }$=