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 450 to the vertical just after release. Then X is nearly

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 450 to the vertical just after release. Then X is nearly (A) 1 × 10-5V  (B) 1 × 10-7V  (C) 1 × 10-9V  (D) 1 × 10-10V

Tardigrade Admin · Accepted Answer

Net force is at 450 from vertical. ∴ qE=mg or (qX/d)=mg (∵ E= (X/d) ) or X= (mgd/q) = ((1.67 ×10-27)(9.8)(10-2)/(1.6 ×10-19)) ≈ 1 × 10-9V

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 $4 5^{0}$ to the vertical just after release. Then X is nearly

$1 \times 1 0^{- 5} V$

$1 \times 1 0^{- 7} V$

$1 \times 1 0^{- 9} V$

$1 \times 1 0^{- 10} V$

Net force is at $4 5^{0}$ from vertical.
$∴ qE = m g$
or $\frac{qX}{d} = m g (∵ E = \frac{X}{d})$
or $X = \frac{m g d}{q}$
$= \frac{( 1.67 \times 1 0 ^{- 27} ) ( 9.8 ) ( 1 0 ^{- 2} )}{( 1.6 \times 1 0 ^{- 19} )}$
$\approx 1 \times 1 0^{- 9} V$

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 450 to the vertical just after release. Then X is nearly

1×10−5V

1×10−7V

1×10−9V

1×10−10V