A charge particle is moving along electric field as shown in figure. What is horizontal displacement of the charge particle when it descends a distance of $y$ meter? Given $Q / m=9.6 \times 10^7 \mathrm{C} / \mathrm{kg}, E=5 \times 10^5 \mathrm{~V} / \mathrm{m}$ , $y=84 \mathrm{~cm}, g=10 \mathrm{~m} / \mathrm{s}^2$

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A charge particle is moving along electric field as shown in figure. What is horizontal displacement of the charge particle when it descends a distance of

y

meter? Given

Q / m=9.6 \times 10^7 \mathrm{C} / \mathrm{kg}, E=5 \times 10^5 \mathrm{~V} / \mathrm{m}

,

y=84 \mathrm{~cm}, g=10 \mathrm{~m} / \mathrm{s}^2

A charge particle is moving along electric field as shown in figure. What is horizontal displacement of the charge

Quantrex Academy · Accepted Answer

Correct answer: A. 4.03 × 10^8 ~m

Suppose particle falls down a distance y in time t so y=(1)/(2) g t^2 ; t= (2 y)/(g) Now for x -axis aligned & x=(1)/(2) a_x t^2=(1)/(2) · (Q E)/(m) · (2 y)/(g) & =(1)/(2) × 9.6 × 10^7 × 5 × 10^5 × 2 × 84 × 10^ -2 10 & =403.2 × 10^ 10 ~m=4.03 × 10^8 ~m aligned

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