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(C) The forces acting on the proton are the gravitational force $mg$ (acting vertically downwards) and the electric force $qE$ (acting horizontally).G

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$1 	imes 10^{-9} \ V$

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(C) The forces acting on the proton are the gravitational force $mg$ (acting vertically downwards) and the electric force $qE$ (acting horizontally).Given that the proton moves at $45^{\circ}$ to the vertical,the magnitudes of the horizontal and vertical forces must be equal:$qE = mg$Here,$q = 1.6 	imes 10^{-19} \ C$,$m = 1.67 	imes 10^{-27} \ kg$,$g = 10 \ m/s^2$,and the electric field $E = \frac{X}{d}$,where $d = 1 \ cm = 0.01 \ m$.Substituting the values:$1.6 	imes 10^{-19} 	imes \frac{X}{0.01} = 1.67 	imes 10^{-27} 	imes 10$$1.6 	imes 10^{-17} 	imes X = 1.67 	imes 10^{-26}$$X = \frac{1.67}{1.6} 	imes 10^{-9} \ V$$X \approx 1 	imes 10^{-9} \ V$

Two large vertical and parallel metal plates having a separation of $1 \ cm$ are connected to a $DC$ voltage source of potential difference $X$. $A$ proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical $JUST$ after release. Then $X$ is nearly

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