Respuesta :
Answer:
[tex]V_b = 600\ V[/tex]
Explanation:
given,
Electric field magnitude, E = 100 V/m
Electric Potential, V_a= 400 V
distance,d= 2 m
Electric potential at Point B = ?
we know.
Δ V = E.d
[tex]V_b - V_a = E.d[/tex]
[tex]V_b - 400 = 100\times 2[/tex]
[tex]V_b= 400 + 200[/tex]
[tex]V_b = 600\ V[/tex]
Hence, the Potential at Point B which is 2 m apart is 600 V
The value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
In terms of potential difference, the electric field can be given as,
[tex]E=\dfrac{\Delta V}{d}[/tex]
Here, (ΔV) is the potential difference and (d) is the distance.
Some region of space contains uniform electric field directed towards south with magnitude 100 V/m.
At point A electric potential is 400 V. The distance of point B directly towards the north from point A is 2 meters. Thus, by the above formula,
[tex]100=\dfrac{ V_b-400}{2}\\V_b=600\rm\; V[/tex]
Hence, the value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.
Learn more about electric field here;
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