Electrostatics Ques 189
- 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
(2012)
(a) $1 \times 10^{-5} \mathrm{~V}$
(b) $1 \times 10^{-7} \mathrm{~V}$
(c) $1 \times 10^{-9} \mathrm{~V}$
(d) $1 \times 10^{-10} \mathrm{~V}$
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Answer:
Correct Answer: 189.(c)
Solution:
Formula:
Coulomb Force Between Two Point Charges:
Net force is at $45^{\circ}$ from vertical.
$$ \begin{aligned} & \therefore \quad q E=m g \text { or } \frac{q X}{d}=m g \quad (\because E=\frac{X}{d}) \\ & \text { or } \\ & X=\frac{m g d}{q}=\frac{\left(1.67 \times 10^{-27}\right)(9.8)\left(10^{-2}\right)}{\left(1.6 \times 10^{-19}\right)} \\ & \approx 1 \times 10^{-9} \mathrm{~V} \end{aligned} $$
BETA
