Question

Electric field intensity is 900 V m⁻¹ at a distance of 2 m from a point charge. It will be 100 V m⁻¹ at a distance?

Answer 1

Explanation:

Electric field intensity due to a point charge is inverse function of distance of point charge from the point where we want to find out the electric field intensity.

Answer 2

Concept: An electric field is that the physical field that surrounds electrically charged particles and exerts force on all different charged particles within the field, either attracting or distasteful them. It additionally refers to the physical field for a system of charged particles.

A measure of the force exerted by one charged body on another. imagined lines of force or field lines originate (by convention) on positive charges and terminate on negative charges.

Given: E1=900 V/m

           r1=2m

           E2=100 V/m

           r2=?

Find: Find the value of r2.

Solution:

E=Kq/$r^{2}$

E1=Kq/$r1^{2}$

E2=Kq/$r2^{2}$

E1/E2=$\frac{\frac{Kq}{r1^{2} } }{\frac{Kq}{r2^{2} } }$

E1/E2=r2^2/r1^2

r2^2=r1^2 XE1/E2

      =4X900/100

      =36

r2=6m

Final answer: Electric field intensity is 100 V m⁻¹ at a distance of 6 m from a point charge.

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