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Mandarinka [93]
2 years ago
8

Which statement is true of the electric field at a distant from the source of charge?

Physics
1 answer:
iVinArrow [24]2 years ago
8 0

Answer:

A negative charge, if free to move in an electric field, will move from a low potential point to a high potential point. To move a positive charge against the electric field, work has to be done by you or a force external to the field.

Explanation:

Mark as Brainliest plz!!!

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An electric field is produced by the very long, uniformly charged rod drawn above. If the strength of the electric field is E1 a
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Answer: hello the complete question is attached below

answer :

r2 = 4r1

Explanation:

Electric field strength = F / q

we will assume the rod has an infinite length

For an infinitely charged rod

E ∝ 1/ r

considering two electric fields E1 and E2 at two different locations as described in the question

E1/E2 = r1/r2 ----- ( 2 )

<u>Calculate for r2 when E2 = E1/4 </u>

back to equation 2

E1 / (E1/4) = r1 / r2

∴ r2 = 4r1

3 0
3 years ago
When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their
AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

F = qvB

Now this causes a centripetal acceleration

F =  = mv^2/r

 so

qvB = mv^2/r ...........(i)

B = mv/(rq)  ...............(ii)

If  accelerating potential V is  same and  then  kinetic energy equals the potential energy difference

\frac{1}{2} mv^2 = Vq

v = \sqrt{(2Vq/m)}      put these value in equation (ii)

B = m\frac{\sqrt{(2Vq/m)} }{rq}  

simplifying we get  

B =m \frac{(\sqrt{ 2Vm/q})}{r}

for same location r will be same in both case

B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}      ..............(iii)

B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}    ..........(iv)

 dividing (iv) and (iii) equation we get

\frac{B_{10}}{B_{7}}   =   \sqrt{\frac{m_{10}}{{m_7}} }

{B_{10}}  =  B_{7}  \sqrt{\frac{m_{10}}{{m_7}} }

B_{10}       = 0.2574T\sqrt{\frac{  (1.663x10^-26}{(1.165x10^-26)}

so on solving we get  

             =0.3075 T

5 0
3 years ago
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