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IRISSAK [1]
3 years ago
4

) At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At wh

at distance from the line will the field strength to be 2000 N/C
Physics
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

0.7071D

Explanation:

Electric field strength is given as:

E = kq/r²

Where k = Coulumbs constant

q = charge

r = distance

When r = D, E = 1000 N/C

=> 1000 = kq/D²

=> kq = 1000D² ______ (1)

When E = 2000 N/C,

2000 = kq/r²

kq = 2000r² ____________ (2)

Since k is a constant and q is the same, equate (1) & (2):

2000r² = 1000D²

=> r² = 1000/2000 D²

r² = 0.5D²

=> r = 0.7071D

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Answer:

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a) As this is a perfectly elastic collision, we can use the formula  v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}+ (\frac{m_{2}-m_{1}}{m_{1}+m_{2}} )v_{2i}, due v_{2i}=0m/s, we obtain v_{2f}=(\frac{2m_{1} }{m_{1}+m_{2}} ) v_{1i}. Then with the data that we know m_{1}=5.21g=0.00521kg, m_{2}=14.8kg and v_{1i}=412m/s, therefore v_{2f}=(\frac{2(0.00521kg) }{0.00521kg+14.8kg} ) 412m/s=0.289m/s or v_{2f}=(0.289\±0.002)m/s adding uncertainty.

b) Now that we know the speed we can use p_{2}=m_{2}*v_{f2} =14.8kg*(0.289\±0.002)m/s=(4.2772\±0.0296)kg*m/s.

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