A parallel plate capacitor is formed from two circular plates, each of radius 1.20 cm, separated by 1.70 mm. The plates are charged to ±±8.70 nC.A proton is shot from the negative plate to the positive. What launch speed must the proton have to just barely reach the positive plate?
Question
A parallel plate capacitor is formed from two circular plates, each of radius 1.20 cm, separated by 1.70 mm. The plates are charged to ±±8.70 nC.A proton is shot from the negative plate to the positive. What launch speed must the proton have to just barely reach the positive plate?
Solution
To solve this problem, we need to equate the kinetic energy of the proton to the potential energy of the capacitor.
Step 1: Calculate the potential energy of the capacitor.
The formula for the potential energy (U) stored in a capacitor is U = 1/2 * C * V^2, where C is the capacitance and V is the voltage.
But first, we need to calculate the capacitance (C) of the capacitor. The formula for the capacitance of a parallel plate capacitor is C = ε0 * (A/d), where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of one of the plates, and d is the separation between the plates.
The area (A) of one of the plates can be calculated using the formula for the area of a circle, A = π * r^2, where r is the radius of the plate.
So, A = π * (1.20 cm)^2 = π * (0.012 m)^2 = 5.44 x 10^-4 m^2.
Then, C = ε0 * (A/d) = 8.85 x 10^-12 F/m * (5.44 x 10^-4 m^2 / 1.70 mm) = 8.85 x 10^-12 F/m * (5.44 x 10^-4 m^2 / 0.0017 m) = 2.79 x 10^-12 F.
The charge (Q) on the capacitor is given as ±8.70 nC, which is ±8.70 x 10^-9 C.
The voltage (V) across the capacitor can be calculated using the formula V = Q/C, so V = (8.70 x 10^-9 C) / (2.79 x 10^-12 F) = 3118 V.
Then, the potential energy (U) of the capacitor is U = 1/2 * C * V^2 = 1/2 * 2.79 x 10^-12 F * (3118 V)^2 = 1.36 x 10^-8 J.
Step 2: Calculate the launch speed of the proton.
The kinetic energy (K) of the proton is equal to the potential energy of the capacitor, so K = U = 1.36 x 10^-8 J.
The formula for the kinetic energy of an object is K = 1/2 * m * v^2, where m is the mass of the object and v is its speed.
The mass (m) of a proton is approximately 1.67 x 10^-27 kg.
So, we can solve for the speed (v) of the proton: v = sqrt((2 * K) / m) = sqrt((2 * 1.36 x 10^-8 J) / 1.67 x 10^-27 kg) = 1.34 x 10^6 m/s.
Therefore, the proton must be launched with a speed of approximately 1.34 x 10^6 m/s to just barely reach the positive plate.
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