Question

ABC is an equilateral triangle. Charges $+q$ are placed at each corner as shown in the figure. The electric intensity at center O will be

• A. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4{\mathrm{\pi \epsilon }}_0$}\right.\right)\left(\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$r$}\right.\right)$
• B. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4{\mathrm{\pi \epsilon }}_0$}\right.\right)\left(\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$r^2$}\right.\right)$
• C.
• D. zero

Answer 1

The correct option is D

zero

Step 1. Given data:

Charges on the vertices of equilateral triangle = $+q$

Step 2. Explanation:

1. Electric intensity $\overrightarrow{E}=\frac{kq}{r^2}$, where $k=\frac{1}{4{\mathrm{\pi \epsilon }}_0}$, ${\epsilon }_0=$the permittivity of free space, $q=$charge, $r=$distance,
2. Since all charges are equal in magnitude and equidistant from point O. Then magnitude of electric field intensity will equal.
3. As Electric field intensity is vector quantity. So direction of vectors of all three will be 120 degree apart from each other.

Step 3. Calculations:

Now ${\overrightarrow{|E}}_A|=|{\overrightarrow{E}}_B|=|{\overrightarrow{E}}_C|=\frac{kq}{r^2}$

Now, Consider ${\overrightarrow{E}}_C$and ${\overrightarrow{E}}_B$, the resultant of these two intensities will be

$$\begin{gathered} E_R=\sqrt{{E^2}_B+{E^2}_C+2E_B{E}_C.Cos120} \\ =\sqrt{{E^2}_B+{E^2}_C-{E^2}_C}AsCos120=-\frac{1}{2},E_B=E_C \\ =E_B \end{gathered}$$

And $E_B$ and $E_A$ are equal in magnitude and opposite in direction.

Hence, the electric intensity at center O will be zero.

Thus, the correct option is option D.