Question

A spherical
conductor of radius
50 cm has a surface
charge density of
8.85 x 10°C/m². The
electric field near the
surface in N/C is​

Answer 1

Therefore the Electric field near the surface is 10⁶ N/C.

Given : A spherical conductor of radius 50 cm has a surface charge density of 8.85 x 10¯⁶ C/m².

To find : The electric field near the surface in N/C.

solution : surface charge density, σ = 8.85 × 10¯⁶ C/m²

so charge on the spherical conductor, Q = 4πr²σ

now electric field due to sphere, E = kQ/r²

= Q/(4πεr²)

= (4πr²σ)/(4πεr²)

= σ/ε

= (8.85 × 10¯⁶)/(8.85 × 10¯¹²)

= 10⁶ N/C

Therefore the Electric field near the surface is 10⁶ N/C