IIT-JEE 2012 · Paper 1 · Physics Q9

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PhysicsSingle correct+3 / -1Question 9 of 60

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° to the vertical JUST after release. Then $X$ is nearly

1\times10^{-5}

1\times10^{-7}

1\times10^{-9}

1\times10^{-10}

Answer and worked solution

Official answer: C

45° to the vertical means the electric force equals the weight:
$eE=mg\implies E=\frac{1.67\times10^{-27}\times10}{1.6\times10^{-19}}\approx1.0\times10^{-7}\ \mathrm{V/m}.$

Voltage. $X=Ed=10^{-7}\times10^{-2}=1\times10^{-9}$ V.

(C)

Topics:Motion in magnetic and electric fields Electric field

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IIT-JEE 2012 · Paper 1, Physics question 9