Question 2: Motion in 2D & Vectors

Question

The square of resultant of two equal electric field vectors is three times their product. Angle between them is

Accepted Answer

Let each field have magnitude $E$.  Then the resultant $R$ of two vectors at angle $	heta$ satisfies
$$ |R|^2 \;=\; E^2 + E^2 + 2E^2\cos	heta \;=\; 2E^2(1+\cos	heta). $$
WeÃ¢ÂÂre told $|R|^2 = 3\,(E\cdot E)=3E^2$.  So
$$ 2E^2(1+\cos	heta)=3E^2 \quad\Longrightarrow\quad 1+\cos	heta=	frac32 \quad\Longrightarrow\quad \cos	heta=	frac12 \;\;\Longrightarrow\;\; 	heta=\frac\pi3. $$
Answer: Option D, $\displaystyle \frac\pi3$.

Answer

$\frac{\pi}{5}$

Answer

$\frac{\pi}{6}$

Answer

$\frac{\pi}{8}$

Answer

$\frac{\pi}{3}$

Motion in 2D & Vectors

The square of resultant of two equal electric field vectors is three times their product. Angle between them is

Choose the correct answer: