A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha
t 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
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