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Please help!

External work equal to 5.0 x 10-7 Jis needed to move a -8.0 nC charge at constant speed to a point at which the potential
is -24 V. What is the potential at the initial point ?

Respuesta :

  • Work=W=5.0×10^-7J
  • Charge=8.0nC=Q
  • Vf=24V

We need potential difference V first

[tex]\\ \sf\longmapsto V=\dfrac{W}{Q}[/tex]

[tex]\\ \sf\longmapsto V=\dfrac{5\times 10^{-7}J}{8\times 10^{-9}C}[/tex]

[tex]\\ \sf\longmapsto V=0.625\times 10^2V[/tex]

[tex]\\ \sf\longmapsto V=62.5V[/tex]

Now

[tex]\\ \sf\longmapsto V=V_f-V_i[/tex]

[tex]\\ \sf\longmapsto V_i=V-V_f[/tex]

[tex]\\ \sf\longmapsto V_i=62.5-24=38.5V[/tex]

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