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sasho [114]
3 years ago
5

Which of the following correctly describe electric field lines?

Physics
2 answers:
Anastasy [175]3 years ago
7 0

Explanation:

The imaginary lines that shows the direction of electric field are called electric field lines. The direction of electric field lines are from positive charge to the negative charge.

Two electric field lines can never ever intersect, if they do so there would be two direction of electric field at that point, which is not possible.

If the electric field lines are very to each other, then it indicates the stronger electric field. The electric field lines is unaffected by the nature of charged particle that created them.

Hence, the correct options are :

(A) They never cross

(B) Field lines that are close together indicate a stronger electric field

(C) They don't affect the charge that created them.

Xelga [282]3 years ago
6 0

-- Electric field lines DO never cross.  <em>(A) </em>

-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B) </em>

-- Electric field lines DO not affect the charge that created them.  <em>(C)</em>

-- Electric field lines DON'T begin on north poles and end on south poles.  North and South "poles" are the way we talk about magnets, not electric charges.

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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solution:

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