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An ion gains 3.9 × 10–18 J of electric potential energy as it moves 22 mm at a 120° angle to a uniform electric field. The electric field strength is 740 N/C.
a) What is the charge of the ion?
b) What is the potential difference between the oil droplet’s starting point and end point in the Sample Problem above?

Respuesta :

onyebuchinnaji

(a) The charge of the ion is determined as 2.76 x 10⁻¹³C.

(b) The potential difference between the oil droplet’s starting point and end point is 16.28 V.

Charge of the ion

The charge of the ion is calculated as follows;

W = Fdsinθ

F = W/dsinθ

F = (3.9 × 10⁻¹²)/(0.022 x sin120)

F = 2.046 x 10⁻¹⁰ N

E = F/q

q = F/E

q = (2.046 x 10⁻¹⁰)/(740)

q = 2.76 x 10⁻¹³C

Potential difference

E = V/d

V = Ed

V = 740 x 0.022

V = 16.28 V

Learn more about potential difference here: https://brainly.com/question/24142403

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