Question

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 ?

Answer 1

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

We need potential difference V first

`\\ \sf\longmapsto V=(W)/(Q)`

`\\ \sf\longmapsto V=(5* 10^(-7)J)/(8* 10^(-9)C)`

`\\ \sf\longmapsto V=0.625* 10^2V`

`\\ \sf\longmapsto V=62.5V`

Now

`\\ \sf\longmapsto V=V_f-V_i`

`\\ \sf\longmapsto V_i=V-V_f`

`\\ \sf\longmapsto V_i=62.5-24=38.5V`