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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)
<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>- 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×
N/C)(2.5m)(1)
- = -3.40×
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
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