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Ymorist [56]
3 years ago
13

A long uniformly charged thread (linear charge density λ= 2.5 C/m) lies along the x axis in the figure.(Figure 1) A small charge

d sphere (Q= -2.3 C) is at the point x=0cm, y=−5.0cm
What is the direction of the electric field at the point x=7.0cm, y=7.0cm?
E
⃗ thread and E Q represent fields due to the long thread and the charge Q, respectively.

What is the magnitude of the electric field at the point x=7.0cm, y=7.0cm

Physics
1 answer:
Kamila [148]3 years ago
5 0

Answer 1) The electric field at distance r from the thread is radial and has magnitude  

E = λ / (2 π ε° r)  

The electric field from the point charge usually is observed to follow coulomb's law:  

E = Q / (4  π ε° r^{2})  

Now, adding the two field vectors:  

E_{thread}  =  {2.5 / (22 π ε° X 0.07 ) ; 0}  

Answer 2) E_{q}  = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))

Adding these two vectors will give the length which is magnitude of the combined field.  

The y-component / x-component gives the tangent of the angle with the positive x-axes.

Please refer the graph and the attachment for better understanding.

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U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

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