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DENIUS [597]
3 years ago
6

Chứng minh V=kq/r từ mối liên hệ giữ E và V

Physics
1 answer:
Marat540 [252]3 years ago
4 0

Answer:V=Aq=KQr

Explanation:

Không biết V bạn kí hiệu ở đây là gì nhỉ? Có phải là điện thế?

Điện thế tại 1 điểm trong điện trường được định nghĩa là công làm vật dịch chuyển từ vị trí đó đến vô cùng. V = A/q

Chứng minh thì được, nhưng chỉ e bạn không có hiểu biết về nguyên hàm, tích phân nên không hiểu.

- Xét tại vị trí cách điện tích Q một đoạn x, khi đó điện tích q sẽ chịu 1 lực: dF=KQ.qx2

Điện tích q dịch chuyển 1 khoảng dx rất nhỏ. Khi đó công do lực điện trường gây ra là:

dA=dF.dx=KQ.qx2dx

Công để dịch chuyển điện tích q từ vị trí r đến vô cùng là:

A=∫∞rdA=∫∞rKQ.qx2dx=KQ.qr−KQ.q∞=KQ.qr

Theo đúng định nghĩa: V=Aq=KQr

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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7
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(a) 1440.5 Hz

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f'=(\frac{v+v_r}{v+v_s})f

where

f is the original frequency

f is the apparent frequency

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v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

v_s is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

v_s = v_A = -28.7 m/s (surface A is the source, which is moving towards the receiver)

v_r = +62.2 m/s (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz

(b) 0.232 m

The wavelength of a wave is given by

\lambda=\frac{v}{f}

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In the reflector frame,

f = 1440.5 Hz

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(c) 1481.2 Hz

Again, we can use the same formula

f'=(\frac{v+v_r}{v+v_s})f

In the source frame (= on surface A), we have

v_s = v_B = -62.2 m/s (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

v_r = +28.7 m/s (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz

(d) 0.225 m

The wavelength of the wave is given by

\lambda=\frac{v}{f}

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m

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