Question

Two point-like charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle as shown in the figure. The electric field intensity at the midpoint $M$ has a magnitude $E_{M}$ . The electric field intensity at the point $P$ (directly above the midpoint) has a magnitude $E_{P}$ . The ratio of these two fields, $E_{M}:E_{P}=27$ . The value of $cos \alpha $ is

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{2}$
• D. $\frac{1}{9}$

Answer 1

Answer: (A)

$E_{M}=\frac{k q \times 2}{r^{2} c o s^{2} \alpha }$

$E_{P}=\frac{2 k q cos \alpha }{r^{2}}$
$\frac{E_{M}}{E_{P}}=\frac{1}{cos^{3} \alpha }=27$
$cos \alpha =\left(\frac{1}{27}\right)^{1 / 3}\Rightarrow cos ⁡ \alpha =1 / 3$