Question

An annular disc has inner and outer radius 30 and 40 respectively. Charge is uniformly distributed. Surface charge density is 10. Find the electric poyrntial at any point distant 30cm from centre of the disc.​

Answer 1

Answer:

see the attachment plz ☺️

Explanation:

mark as brainliast answer ✨

Answer 2

Concept:

The potential due to angular ring can be find out if we subtract the Potential due to air gap from potential due to ring.

Given:

Inner radius of ring is 30

Outer radius of ring is 40

Surface charge density is 10

Find:

Electric potential at 30 from centre

Solution:

Potential due to charge disk

V = (σ/2ε₀) × (√(x²+R²) - x)

Potential due to charge disk at a point P is

Vp = Potential due to whole disk - Potential due to removed portion

Vp = ( (σ/2ε₀) × (√(h²+b²) - h) ) - ( (σ/2ε₀) × (√(h²+a²) - h) )

Vp = (σ/2ε₀) × (√(h²+b²) - h - √(h²+a²) + h )

Vp = (σ/2ε₀) × (√(h²+b²) - √(h²+a²) )

where σ is the charge density, b is the outer radius and a is the inner radius.

Vp = (10/2ε₀) × (√(30²+40²) - √(30²+30²) )

Vp = (10/2ε₀) × (50 - 42.43)

Vp = (5/ε₀) × 7.57

Vp = 37.85/ε₀

So the potential at 30 distant from the centre is 37.85/ε₀.

#SPJ3