Question

Three charges are placed at the vertices of an equilateral triangle of sides 'a'. The force experienced by the charge placed at the vertex A in a direction normal to BC is -

• A. $\frac{Q^{2}}{4\pi \in_{0}a^{2}}$
• B. $–\frac{Q^{2}}{4\pi \in_{0}a^{2}}$
• C. Zero
• D. $\frac{Q^{2}}{2\pi \in_{0}a^{2}}$

Answer 1

Answer: (C)

Zero

Answer 2

F_R = 2F cos q/2

= 2F cos 120/2 = 2F cos 60

= F = $\frac { \mathrm { kQ } ^ { 2 } } { \mathrm { a } ^ { 2 } }$ = $\frac { \mathrm { Q } ^ { 2 } } { 4 \pi \in _ { 0 } \mathrm { a } ^ { 2 } }$

However this force is the resultant force parallel to BC.

\ Force perpendicular to BC is zero.