An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect
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Force = Mass X Acceleration
Mass= Force divided Acceleration Acceleration = Force divided mass
Mass= Force divided Acceleration Acceleration = Force divided mass
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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Explanation:
Given that,
Charge on a spherical drop of water is 43 pC
The potential at its surface is 540 V
(a) The electric potential on the surface is given by :
r is the radius of the drop
(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,
Now the charge on the new drop is 2q. New potential is given by :
Hence, the radius of the drop is and the potential at the surface of the new drop is 856.79 V.