Saved Bookmarks
| 1. |
A spherical conductor of radius50 cm has a surfacecharge density of8.85 x 10°C/m². Theelectric field near thesurface in N/C is |
|
Answer» re 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 = 4πr²σnow 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 |
|