Well I can't see the following physical properties you talked about in the question.
Mass per unit volume ratio is called density.
Power=Energy
72,000/60 = 1200 watts
:)
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)
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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
brainly.com/question/14306881
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Answer:
Explanation:
Relevant Data provided
Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2
Distance which indicates d = 1.50 x 10^-3 m
Voltage which indicates V = 12 V
According to the requirement, the computation of value of its capacitance is shown below:-
=
Therefore for computing the capacitance we simply applied the above formula.