Question

A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.

Answer 1

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

$E = \frac{F}{q} \rightarrow F=Eq$

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

$4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C$

In Second Case,

$KE = q E'd$

$KE = (0.1142)(40)(7)$

$KE = 31.976J$

The total energy change would be subject to,

$\Delta KE = 31.976-4$

$\Delta KE = 27.976J$

Therefore the Kinetic Energy change of the charged object is 27.976J