A mass m carrying a charge q is suspended from a string and placed in a uniform horizontal electric field of intensity E. The angle made by the string with the vertical in the equilibrium position is

Question

A mass m carrying a charge q is suspended from a string and placed in a uniform horizontal electric field of intensity E. The angle made by the string with the vertical in the equilibrium position is (A) θ =tan-1(mg/Eg) (B) θ =tan-1(m/Eg) (C) θ =tan-1(Eq/m) (D) θ =tan-1(Eq/mg)

Tardigrade Admin · Accepted Answer

Correct answer is (d) θ =tan-1(Eq/mg)

Q. A mass m carrying a charge q is suspended from a string and placed in a uniform horizontal electric field of intensity E. The angle made by the string with the vertical in the equilibrium position is

$θ = t a n^{- 1} \frac{m g}{E g}$

$θ = t a n^{- 1} \frac{m}{E g}$

$θ = t a n^{- 1} \frac{Eq}{m}$

$θ = t a n^{- 1} \frac{Eq}{m g}$

Correct answer is (d) $θ = t a n^{- 1} \frac{Eq}{m g}$

Q. A mass m carrying a charge q is suspended from a string and placed in a uniform horizontal electric field of intensity E. The angle made by the string with the vertical in the equilibrium position is

θ=tan−1Egmg​

θ=tan−1Egm​

θ=tan−1mEq​

θ=tan−1mgEq​

Correct answer is (d) θ=tan−1mgEq​

$θ = t a n^{- 1} \frac{m g}{E g}$

$θ = t a n^{- 1} \frac{m}{E g}$

$θ = t a n^{- 1} \frac{Eq}{m}$

$θ = t a n^{- 1} \frac{Eq}{m g}$

Correct answer is (d) $θ = t a n^{- 1} \frac{Eq}{m g}$