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Ne4ueva [31]
3 years ago
11

Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g,

mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2. Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu
Physics
1 answer:
dusya [7]3 years ago
5 0

Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\

Now equate the forces,

T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree

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