contestada

1) An electric field is measured at 3.00 × 10^5 N/C at a distance from a source point charge of 3.00 × 10^–6 C. How far is the point from the source charge?

A) 0.60 m

B) 0.30 m

C) 0.09 m

D) 0.36 m


2) In an electric field of strength 0.50 N/C, a test charge experiences a force of 2.50 × 10^–4 N in the same direction as the field. What is the magnitude of the test charge?

A) +5.00 × 10^–4 C


B) –5.00 × 10^–4 C


C) +2.50 × 10^–4 C


D) –2.50 × 10^–4 C

Respuesta :

rey10mariia
1.a
2. b
i think so those would be the right answers

onyebuchinnaji

(1) The distance of the source charge from the point charge is 0.3 m.

(2) The magnitude of the test charge is +5 x 10⁻⁴ C

Position of the source charge

The position of the source charge is caculated as follows;

[tex]E = \frac{kq}{r^2} \\\\r^2 = \frac{kq}{E} \\\\r = \sqrt{\frac{kq}{E} } \\\\ r= \sqrt{\frac{(9\times 10^9)(3\times 10^{-6})}{3 \times 10^{5}} }\\\\r = 0.3 \ m[/tex]

Thus, the position of the source charge is 0.3 m.

Magnitude of the test charge

The magnitude of the test charge is calculated as follows;

E = F/q

q = F/E

q = (2.5 x 10⁻⁴) / (0.5)

q = 5 x 10⁻⁴ C

Learn more about electric field here: https://brainly.com/question/14372859

Otras preguntas