question_answer

Three charges are placed at the vertices of an equilateral triangle of side a as shown in the following figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is:

A) \[x\]     

B)       \[{{i}_{1}},\]                      

C)  zero                     

D) \[{{i}_{2}},\]

Correct Answer: C

Solution :

From Coulombs law, two stationary point charges \[2\mu F\] and \[{{T}_{2}}>{{T}_{1}}\] repel or attract each other with a force (F) which is directly proportional to the product of charges and inversely proportional to the square of distance (r) between them. \[3q/{{\varepsilon }_{0}}\] The various forces acting at charge +Q are: \[2q/{{\varepsilon }_{0}}\] force of attraction \[q/{{\varepsilon }_{0}}\] \[\beta \] is force of repulsion \[x\] Resolving the force, we have \[y\] along      and \[x\].                                   Two equal and opposite forces which cancel       each other. Hence, net force is zero.