A charged cork ball having mass 1 text{ g} an | vedclass

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(A, B) Mass of the cork ball,$m=1 	ext{ g}=10^{-3} 	ext{ kg}$.Electric field,$E=(3 \hat{i}+5 \hat{j}) 	imes 10^5 	ext{ NC}^{-1}$.Angle,$	heta=37^

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$T=5.55 	imes 10^{-3} 	ext{ N}$

Vedclass · Answer

(A, B) Mass of the cork ball,$m=1 	ext{ g}=10^{-3} 	ext{ kg}$.Electric field,$E=(3 \hat{i}+5 \hat{j}) 	imes 10^5 	ext{ NC}^{-1}$.Angle,$	heta=37^{\circ}$.Force on the cork ball due to the electric field $E$ is $F=qE$.According to the given diagram,resolving all forces in the horizontal and vertical directions:$T \sin 	heta = q E_x \quad \dots (i)$$T \cos 	heta + q E_y = mg \implies T \cos 	heta = mg - q E_y \quad \dots (ii)$Dividing Eq. $(i)$ by Eq. $(ii)$:$	an 	heta = \frac{q E_x}{mg - q E_y}$Substituting the values $(	an 37^{\circ} = 3/4)$:$\frac{3}{4} = \frac{q 	imes 3 	imes 10^5}{10^{-3} 	imes 10 - q 	imes 5 	imes 10^5}$$\frac{3}{4} = \frac{3q 	imes 10^5}{10^{-2} - 5q 	imes 10^5}$$3(10^{-2} - 5q 	imes 10^5) = 12q 	imes 10^5$$0.03 - 15q 	imes 10^5 = 12q 	imes 10^5$$0.03 = 27q 	imes 10^5 \implies q = \frac{0.03}{27 	imes 10^5} = \frac{1}{9} 	imes 10^{-7} \approx 1.11 	imes 10^{-8} 	ext{ C}$.Thus,$q \approx 11 	imes 10^{-9} 	ext{ C}$ or $1.1 	imes 10^{-8} 	ext{ C}$.From Eq. $(i)$:$T \sin 37^{\circ} = q E_x$$T 	imes 0.6 = (1.11 	imes 10^{-8}) 	imes (3 	imes 10^5)$$T 	imes 0.6 = 3.33 	imes 10^{-3}$$T = \frac{3.33 	imes 10^{-3}}{0.6} = 5.55 	imes 10^{-3} 	ext{ N}$.Therefore,options $A$ and $B$ are correct.

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