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 EM . The electric field intensity at the point P (directly above the midpoint) has a magnitude EP . The ratio of these two fields, EM:EP=27 . The value of cos α is

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 EM . The electric field intensity at the point P (directly above the midpoint) has a magnitude EP . The ratio of these two fields, EM:EP=27 . The value of cos α is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-m66k9jhq4qfm.png" /> (A) (1/3) (B) (1/6) (C) (1/2) (D) (1/9)

Tardigrade Admin · Accepted Answer

EM=(k q × 2/r2 c o s2 α ) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-m66k9jhqpiwz.png" /> EP=(2 k q cos α /r2) (EM/EP)=(1/cos3 α )=27 cos α =((1/27))1 / 3⇒ cos â¡ α =1 / 3

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{9}$

$E_{M} = \frac{k q \times 2}{r ^{2} co s ^{2} α}$

$E_{P} = \frac{2 k q cos α}{r ^{2}}$
$\frac{E _{M}}{E _{P}} = \frac{1}{co s ^{3} α} = 27$
$cos α = (\frac{1}{27})^{1/3} \Rightarrow cos α = 1/3$

31​

61​

21​

91​

EM​=r2cos2αkq×2​ EP​=r22kqcosα​ EP​EM​​=cos3α1​=27 cosα=(271​)1/3⇒cos⁡α=1/3

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{9}$

$E_{M} = \frac{k q \times 2}{r ^{2} co s ^{2} α}$

$E_{P} = \frac{2 k q cos α}{r ^{2}}$
$\frac{E _{M}}{E _{P}} = \frac{1}{co s ^{3} α} = 27$
$cos α = (\frac{1}{27})^{1/3} \Rightarrow cos α = 1/3$