Question

The two charges in the figure below are separated by d = 2.50 cm.

(Let q₁ = −16.0 nC and q₂ = 25.5 nC.)

Two charges and two points lie along the perimeter of an equilateral triangle with side length d (interior angle 60.0°).

• Negative charge q₁ is at the bottom left vertex.
• Positive charge q₂ is at the bottom right vertex.
• Point A is at the top vertex.
• Point B is on the bottom edge, midway between q₁ and q₂.

(a) Find the electric potential at point A.
kV

(b) Find the electric potential at point B, which is halfway between the charges.
kV

Answer 1

poitential at point a is given by Va = V, + V₂ Va = kali + ke 7 / 2 (2, +22) given q = -16.onc Es-snc = 0.025 meter d = 2-50 cm 0.025 (b) va = 9X10% (-16.0 +25.5) x10? [va = +3420 volt] = 3.42 kV Potatiol of point B is given by Vo = v tre V = k g t I Vo = 34 (9, +22) Vo = 23.760% (-16.0 Svo = 6-84 kv] 16.0 +25.5) X10 troe

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