Physics College

what is elecric field line​

Answers

Answer 1

A vector field and a starting position inside the field define the locus as a "field line." We have electric field lines for the electric fields.

What do you mean by electric field line?

In general, an electric field line is a curve drawn with each point's tangent pointing in the direction of that point's net field. An arrow on the curve is obviously required to designate the direction of the electric field from the two potential directions provided by a tangent to the curve. A field line is a space curve or a three-dimensional curve.The vector gets shorter as you move away from the origin and always points radially outward because the electric field varies as the inverse of the square of the distance that points from the charge.The density of the lines indicates the size of the field. This indicates that the location with a high density of field lines has a stronger electric field due to the charged substance. The electric field is weaker in the area where the density of these lines is low.The electric field can be depicted by joining these vectors to form a line.The direction of an imaginary line called an electric field line at any place must match the direction of the field there.In general, a field line is a curve drawn so that each point's tangent points in the direction of the net field.

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Related Questions

A block and tackle pulley system has 3pulley wheels in the lower movable block. Determine the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​

Answers

A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort

How to calculate  the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​?

VR =3

VR = n=3

Efficiency of the system = 80%

Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4

Man can lift load with effort =350N

Thus,

Load= MA× effort = 2.4×350 = 840 N.

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