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When a [tex]-3.72 \cdot 10^{-4} \, C[/tex] charge moves from point [tex]A[/tex] to point [tex]B[/tex], its electric potential energy increases by [tex]0.218 \, J[/tex].

What is the potential difference between [tex]A[/tex] and [tex]B[/tex]? Include the correct sign, + or -.

(Unit [tex]= V[/tex])



Answer :

GinnyAnswer
To determine the potential difference between points [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we start by using the relationship between electric potential difference (voltage), charge, and change in electric potential energy. The formula for electric potential difference [tex]\( V \)[/tex] is given by:

[tex]\[ V = \frac{\Delta E}{q} \][/tex]

where:
- [tex]\( \Delta E \)[/tex] is the change in electric potential energy (in Joules),
- [tex]\( q \)[/tex] is the charge (in Coulombs).

We are given:
- The charge [tex]\( q = -3.72 \times 10^{-4} \, \text{C} \)[/tex],
- The change in electric potential energy [tex]\( \Delta E = 0.218 \, \text{J} \)[/tex].

Substituting the given values into the formula, we get:

[tex]\[ V = \frac{0.218 \, \text{J}}{-3.72 \times 10^{-4} \, \text{C}} \][/tex]

Performing the division, we find:

[tex]\[ V = -586.0215053763441 \, \text{V} \][/tex]

So the potential difference between point [tex]\( A \)[/tex] and point [tex]\( B \)[/tex] is:

[tex]\[ -586.0215053763441 \, \text{V} \][/tex]

The negative sign indicates that the electric potential at point [tex]\( B \)[/tex] is lower than the electric potential at point [tex]\( A \)[/tex].

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