Respuesta :
Explanation:
As per the problem,
[tex]\Delta U = (V_{B} - V_{A})(-q) > 0[/tex]
When q > 0 then -q is a negative charge . Since, change in potential energy ([tex]\Delta U[/tex]) increases.
or, [tex](V_{A} - V_{B})q[/tex] > 0
or, [tex]V_{A} > V_{B}[/tex]
Therefore, both positive and negative charge will move from [tex]V_{A}[/tex] to [tex]V_{B}[/tex] and as [tex]V_{B} < V_{A}[/tex] so both of them move through a negative potential difference.
Thus, we can conclude that the true statements are as follows.
- The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
- The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
The correct option is :
(B) The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
(C) The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
Electric potential energy:
ΔU = -qΔV, where q is the charge and ΔV is the potential difference
now, [tex]\Delta V=V_A-V_B[/tex]
According to the question, an object with a negative charge is moved from point A to point B through an external electrical field, it gains electrical potential energy.
since q is negative then, -q is +ve
also ΔU is given +ve,
which means
[tex]V_A-V_B>0\\\\V_A>V_B[/tex]
[tex]V_B-V_A<0[/tex]
- We come to the same conclusion if we take the case of positive charge; since in that case q is +ve, so -q will be -ve, and ΔU is also -ve, so ΔV must be +ve, which gives the same result.
- Therefore from the options given we can say that The positively and negatively charged object moves through a negative potential difference between A and B.
Learn more about electric potential energy:
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