Question

A point charge of 0.33×10^-8C is placed in a medium of relative permittivity of 5 . Calculate electric field intensity at a point 10cm from the charge

Answer 1

electric field intensity at a point 10cm from the charge is 594 N/C.

It is given that a point charge of 0.33 × 10^-8 C is placed in a medium of relative permittivity is 5.

we have to calculate electric field intensity at a point 10cm from the charge.

using Coulomb's law,

E = Kq/μr²

where μ is relatively permittivity i.e., μ = 5

r = 10cm = 0.1 m , K = 9 × 10^9 Nm²/C² and q = 0.33 × 10^-8 C

so, E = (9 × 10^9 × 0.33 × 10^-8)/{5 × (0.1)²}

= 29.7/0.05

= 594 N/C

therefore, electric field intensity at a point 10cm from the charge is 594 N/C.

Answer 2

Answer:

Explanation:

E = Kq/μr²

where μ is relatively permittivity i.e., μ = 5

r = 10cm = 0.1 m , K = 9 × 10^9 Nm²/C² and q = 0.33 × 10^-8 C

so, E = (9 × 10^9 × 0.33 × 10^-8)/{5 × (0.1)²}

= 29.7/0.05

= 594 N/C