Question

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x=2.50m? (a) 3.40 × 10⁻¹⁶ J(b) -3.40 × 10⁻¹⁶ J(c) 2.50 × 10⁻¹⁶ J (d) -2.50 × 10⁻¹⁶ J(e) -1.60 × 10⁻¹⁹ J

Answer 1

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

How is the change in electric potential energy of the proton-field system calculated?

• Work done on the proton =Negative of the change in the electric potential energy of the proton field
• In the given case, W = -qΔV
• -W = qΔV
• = qEcosθ
• Therefore, work done on the proton = -e(8.50×$10^2$ N/C)(2.5m)(1)
• = -3.40×$10^-^1^6$ J
• Any change in the potential energy indicates the work done by the proton.
• Therefore the positive sign shows that the potential energy increases when the proton does the work.
• The negative sign shows that the potential energy decreases when the proton does the work.

To learn more about electric potential energy, refer

brainly.com/question/14306881

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