JEE Advanced · Physics · 5. Laws of Motion
Two large vertical and parallel metal plates having a separation of \(1 \mathrm{~cm}\) 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
- A \(1 \times 10^{-5} \mathrm{~V}\)
- B \(1 \times 10^{-7} V\)
- C \(1 \times 10^{-9} V\)
- D \(1 \times 10^{-10} V\)
Answer & Solution
Correct Answer
(C) \(1 \times 10^{-9} V\)
Step-by-step Solution
Detailed explanation
According to the question, a proton is released at rest mid-way between the two plates and is found to move at \(45^{\circ}\), so net force is at \(45^{\circ}\) from vertical and two forces acting on the proton just after the release are as shown in the figure.
\(q E=m g\)
\(\therefore \quad V=\frac{m g d}{q}=\frac{1.67 \times 10^{-27} \times 10 \times 10^{-2}}{1.6 \times 10^{-19}}=10^{-9} \mathrm{~V}\)
\(q E=m g\)
\(\therefore \quad V=\frac{m g d}{q}=\frac{1.67 \times 10^{-27} \times 10 \times 10^{-2}}{1.6 \times 10^{-19}}=10^{-9} \mathrm{~V}\)
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