Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference . A proton is released at rest midway between the two plates. It is found to move at to the vertical JUST after release.

Question

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 [2012]

Smart Achievers · Accepted Answer

(3)

According to the question, a proton is released at rest midway between the two plates and is found to move at $45 °$ ,

Therefore, the net force is at $45 °$ from the vertical, and the two forces acting on the proton just after release are as shown in the figure.

$q E = m g$

$∴$ $q (\frac{V}{d}) = m g$

$∴$ $V = \frac{m g d}{q}$ $= \frac{1.67 \times 10^{- 27} \times 10 \times 10^{- 2}}{1.6 \times 10^{- 19}}$ $\approx 10^{- 9} V$

Hence, $X \approx 1 \times 10^{- 9} V$

Solution

Q.
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 [2012]

1 $1 \times 10^{- 5} V$

2 $1 \times 10^{- 7} V$

3 $1 \times 10^{- 9} V$

4 $1 \times 10^{- 10} V$

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Solution

Q. 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 [2012]

1 1×10-5 V

2 1×10-7 V

3 1×10-9 V