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the electric field strength is 5.40×104 n/c inside a parallel-plate capacitor with a 2.10 mm spacing. a proton is released from rest at the positive plate.

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Final answer:

The question involves calculating the initial acceleration of a proton in a specific electric field using the charge of the proton, the electric field strength, and Newton's second law.

Step-by-step explanation:

The subject question asks to calculate the initial acceleration of a proton released from rest in an electric field inside a parallel-plate capacitor. The proton experiences a force due to the electric field, which can be calculated using the formula F = qE, where F is the force on the proton, q is the charge of the proton, and E is the electric field strength. The charge of a proton is approximately 1.60×10-19 coulombs, and the given electric field strength is 5.40×104 N/C. Once the force is calculated, Newton's second law (F = ma) is used to find the acceleration, a, with m being the mass of the proton (approximately 1.67×10-27 kg).

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