Question

Charges +2q, +q and +q are placed at the corners A, B and C of an equilateral triangle ABC. If E is the electric field at the circumcentre O of the triangle, due to the charge +q, then the magnitude and direction of the resultant electric field at O is
(a) E along AO
(b) 2E along AO
(c) E along BO
(d) E along CO
(e) Zero

Answer 1

As a first step, draw a diagram representing all the given fields. If you find it necessary, resolve the electric fields into its components and also see whether you could cancel any of these components. After doing all these, find the resultant of the remaining components and hence find the answer.

Complete step-by-step answer :
We are given an equilateral triangle ABC with circumcentre O. We also have charges +2q, +q and +q placed at corners A, B and C respectively.
We are also given the electric field due to charge +q as E at the circumcentre O. So, we have two electric fields E along BO and CO due to charge +q kept at corners B and C respectively.
Let us recall that, the electric field due to charge q at a distance r is given by,
$E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{r}^{2}}}$
So, if the electric field due to charge +q is E at a point, then, electric field due to charge +2q kept at the same distance from the point should be 2E. Therefore, we have an electric field 2E along AO due the charge +2q kept at corner A. Let us represent all these electric fields in a diagram.

From the geometry of an equilateral triangle, the angle made by the electric field vectors with AO is 60°. If we resolve the two electric fields due to the two +q-charges into its vertical and horizontal components, we see that the horizontal components get cancelled due to each other and only vertical cosine components of each field remain. That is, along OA we have,
$E\cos 60{}^\circ +E\cos 60{}^\circ =2E\left( \dfrac{1}{2} \right)=E$
Now, we have to find the resultant of 2E along AO and E along OA to find the net electric field O due to the given arrangement.
So, net electric field at O will be,
$2E-E=E$, directed along AO.
Hence, the answer to the given question is option A.

Note : Remember that, if the electric field due to q charge is E then, that due to 2q should be 2E from the expression of electric field. In questions like this, where we are asked to find the resultant, make sure that you resolve the vectors into its components wherever necessary. Also, drawing a diagram as a first step gives you an idea on how to approach this problem.