Respuesta :
Answer:
The potential difference between points A and B is 278.95 volts.
The potential difference between points B and C is -642.10 volts.
The potential difference between points A and C is -363.15 volts.
Explanation:
Given :
Charge of the particle, q = 19 nC = 19 x 10⁻⁹ C
Work is done to move a charge from point A to B, W₁ = 5.3 μJ
Work done to move a charge from point B to C, W₂ = -12.2 μJ
Let V₁ be the potential difference between point A and B, V₂ be the potential difference between point B and C and V₃ be the potential difference between point A and C.
The relation between work done and potential difference is:
W = qV
V = W/q ....(1)
Using equation (1), the potential difference between points A and B is:
[tex]V_{1}=\frac{W_{1} }{q}[/tex]
Substitute the suitable values in the above equation.
[tex]V_{1} =\frac{5.3\times10^{-6} }{19\times10^{-9} }[/tex]
V₁ = 278.95 V
Using equation (1), the potential difference between points B and C is:
[tex]V_{2}=\frac{W_{2} }{q}[/tex]
Substitute the suitable values in the above equation.
[tex]V_{2} =\frac{-12.2\times10^{-6} }{19\times10^{-9} }[/tex]
V₂ = -642.10 V
The potential difference between points A and C is:
V₃ = V₁ + V₂
V₃ = 278.95 - 642.10
V₃ = -363.15 V
